Линейные и нелинейные цепи (на английском языке). Linear and nonlinear circuits 9785999401182

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K.K. Ким

ЛИНЕЙНЫЕ И НЕЛИНЕЙНЫЕ ЦЕПИ Учебное пособие

Москва 2011

K.K. Kim

LINEAR AND NONLINEAR CIRCUITS CIR CUITS Tutorial

Moscow 2011

УДК 656.423 ББК 39.232 К40 Автор — заведующий кафедрой «Т «Теоретические еоретические основы электротехники» Петербургского государственного университета путей сообщения, член Ака демии электротехнических на наук ук Российской Федерации, член Европейской академии естественных на наук, ук, член Института инженеров по электротехнике и радиоэлектронике (США), член НьюЙоркской академии нау наук, к, др техн. наук, проф. K.K. Kим Рецензент — проф. кафедры «Т «Теоретические еоретические основы электротехники» СанктПетербургского государственного политехнического университета, др техн. нау наукк М.А. Шакиров

Kim K.K. Linear and Nonlinear Circuits: Tutorial / K.K. Kim. — Moscow: К40 FSET «Education and Methodics Center of Railwa y Education», 2011. — 200 p. — На англ. яз. (Ким К.К. Линейные и нелинейные цепи: учеб. пособие. — M.: ФГО ФГОУ У «У «Учебнометодический чебнометодический центр по образованию на железнодорожном транспорте», 2011. — 200 с.) ISBN 978 5 9994 0118 2 Учебное пособие, написанное на английском языке, составлено в соответствии с программой курса «Т «Теоретические еоретические основы электротехники» и предназначено для студентов электромеханических и электротехнических специальностей. В пособии рассматриваются аспекты синтеза электрических цепей с сосредоточенными параметрами, теории электрических цепей с распре деленными параметрами и некоторые вопросы теории нелинейных элек трических и магнитных цепей. Данное учебное пособие является про должением книги К.К. Кима «Линейные электрические цепи», изданной в 2006 году году.. УДК 656.423 ББК 39.232

ISBN 978 5 9994 0118 2

© Ким К.К., 2011 © ФГО ФГОУ У «Учебнометодический центр по образованию на железнодорожном транспорте», 2011 © Оформление. ООО «ПиарПресс», 2010

The course «Theoretical Fundamentals of Electrical Engineering» con sists of the theory of electrical and magnetic circuits and the theory of elec tromagnetic fields. This course (according to the curriculum) is taught during three semesters. The synthesis of electric circuits with lumped parameters, the aspects of the theory of electrical circuits with distributed parameters and som some e problems of the theory of nonlinear electrical and magnetic circuits are considered in this tutorial. The tutorial is done for students of electrical, electronic, control, com munication, information, instrumentation and computer engineering spe cialties. There are calculation hometasks which are giv given en to the students at the end of the tutorial. There is also a list of the recommended literature for the students.

1. SYNTHESIS OF LINEAR PASSIVE ELECTRI ELECTRICA CAL L CIRCUITS WITH LUMP LUMPED ED PARAMETERS The problem of analysis is to calculate currents, voltages in the given circuit. Earlier we considered only this problem. The problem of synthesis is to construct an electrical circuit, in which processes will flow by the giv given en law law.. Let’s set up a task to construct a circuit (a tw two o pole network) with lin ear elements. This circuit should hav have e the required characteristics. The in put value x1(t) and its operator image 1X(p) (x1(t) is either the current or the voltage) are known. The required law of changing the output value 2(t) x is giv given, en, its operating image is2(Xp). Thus we have a formula of the transfer X ()p have e this transfer func function Kp() = 2 . The created circuit should hav X ()p 1 tion. Hence, the problem is to construct an electrical circuit by the giv given en transfer function K(p) or by the giv given en frequency characteristic K(jω). The input impedance and admittance of a tw two o pole network can be considered as special cases of the transfer function. For this case we should take one of the tw two o values (either the current or the voltage) as the input value and the other one as the output value. The input and transfer functions of the cir cuit are simply named the functions of a circuit and denoted by F(p). F(p) are rational fractions of the complex frequency p = σ + jω for line ar circuits with lumped parameters. The properties of F(p) and the proper ties of electrical circuits, described by them, are characterized unambigu ously by the positions of zeros and the poles of F(p). F(p) can be given as frequency characteristics. When we considered electrical circuits, we saw that various electrical circuits could hav have e identical functions of a circuit by form, for example, differentiating circuits, consisting of R, L and R, C integrating circuits, consisting of the same elements. Hence we see that the same problem of synthesis can hav have e various solutions. Such cases are also possible when there is no solution of a concrete problem of synthesis with the help of linear passiv passive e elements ( R, L, C). 4

In this connection the following main questions of synthesis of a circuit by the giv given en function of the circuit F(p) appear: 1. Rev Revealing ealing opportunities of constructing a circuit by the giv given en F(p) with the help of usual elements as R, L, C. 2. Working out the method of concrete realization of the giv given en F(p) as an electrical circuit.

1.1. The properties of input functions of passiv passive e electrical circuits (tw two o pole netw networks) orks) The basic properties of input operational impedances Z(p) and admit tances Y(p) of passiv passive e electrical circuits (i.e. the circuits, consisting of R, L, C) are as follows: 1. Z(p) and Y(p) are real values at the real p (p ). The polynomials in the numerator and denominator of the functions Z(p) and Y(p) are real, since they are formed by the sums, differences, products and quotients of the real parameters R, L, C of the subcircuits. 2. The poles and zeros of the functions Z(p) and Y(p) are situated only in the left half plane p j or on the axis of the imaginary numbers, i.e. 0 and in this casek 0 the poles and zeros are simple. Here all the k factors of the polynomials of p, situated in the numerator and the deno denom m inator,, are positive. inator Let’s factorize the polynomial:

annpnn a

p

1

10

There are multipliers: pp

p

2 kk

2

a ii kk

an pp pp

1

pp

n

p

k

n 1

jk

pp 1 . p

k

i for each pair of the complex rootspki and pp kk 1

jk

and

the multipliers as p –p p – i for the real roots p i i . From here we can see that if all k 0 and i 0, the multipliers, into which the polynomial is fractorized, don’t contain negativ negative e numbers, hence, the factors n, an–1 …, a0 are positiv positive e numbers. 3. The real part of the functions Z(p) and Y(p) is not negative (i.e. it is positive positiv e or equal to zero): Re[Z(p)] 0 and Re[ Re[Y Y(p)] 0, if 0. Let 0 be, i.e. p j . In this case Z( j ) and Y( j ) are usual co complex mplex impedance and admittance. At the presence of an activ active e resistance ev even en in one branch of a two pole netw network, ork, the activ active e pow power er is positive at the input of 5

the tw two o pole network and, hence, the activ active e resistance and active conduc tance of the whole tw two o pole netw network ork are positive too, i.e. Re[ Re[Z Z( jω)] ≥ 0 and Re[Y( jω)] ≥ 0. When there are only wattless elements in the circuit, Re[Z( jω)] = 0. Let’s show that Re Re[[Z( p)] > 0 even for a “pure” wattless circuit if σ > 0. For the circuit shown in Fig. 1.1, we have

Zp() =+pL Fig. 1.1. The circuit cor responding to Re[ Z( p)]

11 += ()σ+ jCω

1 = ()σ + j ω L + pC

σLj + ωL +

σC + j ωC

. (1.1)

This equation coincides with the equation for the complex impedance of the circuit by form: 1 . Zj() ω= R + jLω + Gj+ ωC

This circuit is giv given en in Fig. 1.2.

Fig. 1.2. The circuit corresponding toZp() = σ L + jLω +

1 σ+ Cj

ωC

The complex impedance at R > 0 and G > 0 has a positiv positive e real part, just as the real part of the operating impedance (1.1) is positiv positive e at σ > 0. For any complex circuit consisting only of wattless elements, by analogy analogy,, the circuit containing activ active e elements can be constructed too. In this case the active activ e resistance iR= σ Li is connected with the inductance iLin series in and the conductance G i = σ Ci is connected with each capacitance i C parallel. Thus the operating input impedance Z(p) of the wattless circuit at p = σ + j ω and σ > 0 will be similar by form to the co complex mplex impedance Z(jω ) of the whole circuit with the added activ active e elements.

6

The functions, having the abov above e indicated properties, are named posi tive real functions. From the abo above ve stated it follows that in order the rational fraction:

Fp

nn ap a nn

p 10

mm bp b mm

10

p

1

...

1

a

...

b

Gp Qp

should be an operating input function and could be realized as a concrete electrical circuit, it must satisfy the abo abovve listed requirements, i.e. G(p) and Q(p) should hav have e zeros in the left half plane or on the axis of the imag positive e real numbers inary numbers, all the factors kaand bk should be positiv and at last the inequality Re[F(p)] 0 should be carried out at Re Re((p) 0. Besides the degrees of the polynomials («n» and «m») should not differ from each other more than per unit.

1.2. The representation of input functions of tw two o pole netw networks orks as simple fractions (Foster’s method) It is possible to represent an input function F(p), which is a rational fraction, as the sum of the terms Ap + A0 and simple fractions: Gp

Fp

Ap

Q P

A0

A1 pp

Am pp

1

, m

where p1,…, pm are the roots of Q(p) 0. In this case A 0, if n m +1. A 0 and A 0 at n m . When n m –1, the factors AA0 0. At an Fp n m +1, A , the factors A1,…, Am are calculated af p bm p

have e been separated by the known method. ter A p + A0 hav Let’s consider a case, when the roots of Q(p) 0 are imaginary and real numbers. The imaginary roots should be conjugated in pairs. Let k p j k and *A pk+1 –j k be. In this case kA kA+ jA k and Ak+1 k kA– jA k should be. Therefore Ap j Ak p j k AA kk kk

pp kk

1

pp

1

pj

kk

pj 7

∗∗

++ k p AA kk ==

AA−

k

jωk

′′ 22 Ap kk

22+ ω pp kk

22+ ω

−

A′ ωk p22+ ωk

.

If Re[F(p)] ≥ 0 at σ ≥ 0, then A′k = 0. In fact, let p = σ → 0 be, then under ′ ≤ 0. the condition Re Re[[F(p)] ≥ 0 it follows that –2A′k ωk ≥ 0 and, hence, A k 2 2 2 2 Let p = jω and ω > kωbe now now,, i.e. p+ ωk = –ω + ω k < 0. Then from

′ω 2Akk

Re[F(p)] ≥ 0 it follows that −≥ −ω22+ ω

0 , i.e. A′k > 0. Both the inequal

k

ities for A′k are satisfied only under ′kA= 0. Therefore AA 2 Ak′ p Bk p +1 kk += = , 22+ ω 22+ ω pp−− kk pp +1 pp kk

Bk = 2A′k

is the real number number.. If the root ip= σi = –δi is real, the corre AA sponding fraction has the following form: ii = . I.e. at the pre pp− ii p+ δ where

sence only of real and imaginary roots we hav have e Fp() =

Gp() Qp()

Bp 1 = Ap + +++ A ∞ 0 22+ ω pp 13 AA−1 ++ mm pp+ δmm −1

+δ

.

Bp 3 22+ ω

... +

(1.2)

1.3. The realization of input functions of a tw two o pole netw network ork with real and imaginary roots of the denominato r (Foster’s method) Let F(p) be the input operating impedance of a tw two o pole netw network ork in the equation (1.2), F(p) = Z(p). Further let’s assume that all the factors in (1.2) are real and positiv positive. e. We shall consider how separate terms in (1.2) can be realized. 8

The term A∞p is realized with the help of a coil with the inductance L ∞ = A∞, since the operating reactance for it is equal to∞pL. The term A0 is realized by a resistor with an activ active e resistance 0 = A0). 0(RR Bp 1 = The term 22k is realized with the help of the subcircuit, ω2 p + ωkk p + Bk pB

()

k

shown in Fig. 1.3.

Fig. 1.3. The circuit corresponding to

The term

Ai p + δi

=

1 p δi + AA ii

1 ω2 p + k Bp B kk

is realized by the subcircuit shown in Fig. 1.4.

Fig. 1.4. The circuit corresponding to

1

p δi + AA ii

9

Zp() =+Ap A0 + Thus for a concrete case it is possible to write ∞ Bp A3 . This equation corresponds to the circuit shown in ++ 1 22+ ω + δ p p 3 1 Fig. 1.5.

L ∞ = A∞

R0 = A0

Bp A + 3 Fig. 1.5. The circuit corresponding toZp() =+A∞ p A0 + 221 p + ω1 p + δ3

Let F(p) represent the input operating admittance in the equation (1.2), F(p) = Y(p). Then the term A ∞p is realized with the help of the capacitor∞C (C∞ = A∞), since the operating admittance for it is ∞Cp. active e conductance The term A0 is realized by the subcircuit with the activ G0 (C0 = A0). The term

Bp k p22+ ωkk

=

1 p + Bp kk

ω2

is realized by the subcir

B

cuit shown in Fig. 1.6, since its operating admittanceYp isk () =

Bp 1 = Fig. 1.6. The circuit corresponding to 22k ω2 p + ωkk p + Bp B kk

10

1 1 pLk + pCk

.

The term

Ai

=

1

is realized by the subcircuit (Fig. 1.7), since p δi + AA ii its operating admittance is calculated by the following equation 1 = () Yp . i pLR+ p + δi

ii

Ai 1 = Fig. 1.7. The circuit corresponding to p + δi p δi + Aii A

Thus it is possible to writeYp()

=+A∞ p A + 0

Bp 1

+

A3

for a con p22+ ω1 p + δ3 crete case and in accordance with this equation we hav have e the following cir cuit (Fig. 1.8). It is possible the realization with the help of the equation (1.2) in some special cases at the negativ negative e magnitude i of if AA0 is rather a large value. AA A0 +=+ ii pp+ δii

C∞ = A∞

G0 = A0

00

A δ+ A i

+δ

p + δi

p

L 1 = 1/B1 C1 = B1/ω 21

.

L3 = 1/A3

R3 = δ3/A 3

Bp A + 3 Fig. 1.8. The circuit corresponding toYp() =+A∞ p A0 + 221 p + ω1 p + δ3

11

If A 0δi +A i > 0, the term

δ+ AA 0 ii p + δi

is realized as the circuit shown in

Fig. 1.9, or as the circuit shown in Fig. 1.10.

Fig. 1.9. The circuit corresponding to

Fig. 1.10. The circuit corresponding to

δ+ AA 0 ii

δ+ AA 0 ii

p+ δ

p+ δ

i

The fraction

i

Ap 0 p + δi

=

1 1 + AA 00

is realized by the subcircuit (Fig. 1.11)

δ

i

p

in case F(p) = Z (p), as for itZp() =

1 1 G0 + pL 0

. In case of F(p) = Y(p) this

fraction is realized by the following subcircuit (Fig. 1.12).

Fig. 1.11. The circuit corresponding to

Ap 0 p + δi

12

=

1 δ 1 + i A00 Ap

Fig. 1.12. The circuit corresponding to

1 G0 +

1 pL0

Let’s consider an example of realization of Z(p). Let’s express the resis 1 and frequency ω in relativ relative e units in order the factors of ,ω , tancesRL ωC polynomials should be small. Let Zp()

p53++68pp == 42++ pp 43

p53++68pp

()pp 13(

)

22++

.

= ± , p = ± j 3. The denominator has only imaginary rootspj1,2 ,2 3,4 Therefore Zp() =+Ap ∞

Bp 1 22+ ω pp 13

+

Bp 3 22+ ω

22= ω = , ω13 1, 3.

Besides A 0 = 0 (we are convinced in it by substituting p = 0 into the equation for Z(p)). == AB ∞

Zp()

1;

pp ==

1

=

Zp()

()p + ω 22

1

p=∞

p22= − ω1

42++ pp 68 2

p +3

p2 = −1

By analogy we get Zp() p22+ ω3 B3 == p

()

− +8 3 16 = . 22 42++ pp 68

p22= − ω3

p2 + 1

p2 = −3

1 = . 2

Thus the parameters of the circuit (Fig. 1.13) realizing the giv given en B1 31 B3 1 2 == = = = = = ; 1 ; ; C ; L function Z ( p ) are equalLL ∞ 11 3 22 23 B1 6 ωω 13 1 C3 == 2. B3

Fig. 1.13. The circuit corresponding toZp()

=

53++ pp 68

p

42++

pp 43

13

1 . Zp() A Bp B −p = 1 + 35 + . 2+ 2+ p p 2 p 4

Yp() = Now let’s consider the operating admittance 42++ pp 43

42++ pp 43

Yp() == 53++ pp 68

p

()

pp

22++

(

)

24 p

In this case A∞ = A0 = 0, since Y(p) is the proper fraction. == AY 13

()p

p

p=0

Yp() 31 ; B= 84

()p + 2 2

= ;

p p2 = −2

Yp() B5 ==

()p + 4 2

3 . 8

p p2 = −4

18 == LL ; We hav have e a circuit shown in Fig. 1.14. In this circuit 13 A 3 1 BB 11 1 8 3 == = 35 = ; == = = . 4; CL ; C 35 22 83 5 BB 32 ωω 35

=

35

Fig. 1.14. The circuit corresponding toYp() =

42++ pp 43 53++ pp 68

p

1.4. The realization of input functions of a tw two o pole netw network ork having ha ving only imaginary roots of the denominator If the denominator of the input functions Z(p) and Y(p) has only imag inary roots, the corresponding circuit consists only of wattless elements. Therefore in the equation (1.2) the terms 0Aand Ai /(p + δ i ) should be 14

absent, since at realizating them the activ active e resistances should be used. Due to the abo abov ve statement of F(p) should be written as: Fp

Gp Qp

Ap

Bp 3

Bp 1 22 pp

...

22

p A k

13

Bk

∑

22 1,3 , 3 .5 . .. p

. k

From this example we can see that, if allk 0, Q(p) will be a complete polynomial of the ev even en degrees of p, i.e. the polyno polynomial mial containing all (with out missing) the ev even en exponents from «0» up to «m». In this case G(p) will be a complete polyno polynomial mial of odd degrees. Taking it into consideration, we shall write Fp

Gp Qp

mm11 ap ap mm11 mm bp b mm

a1 p

p 20

2

...

where m is an

b

even ev en number number.. The value of p (p 0) is zero of F(p). If one of the roots of Q(p) is equal to zero, b0 0. In this case dividing the numerator and the deno denominator minator by p, we shall get G(p), which is the polynomial of ev even en degrees and Q(p), which is the polynomial of odd degrees. In order to realize F( p) as an electrical circuit consisting of wattless elements it is necessary that F(p) should satisfied the abov above e mentioned re quirements, namely: the degrees of the polynomials G(p) and Q(p) should differ from each other per unit; the zeros and poles of F(p) should alter nate, i.e.

Fp

Gp Qp

22

22

ap p mm12 bp mm 0.

p

22

123

22

13

p

..

4

... p22 ,

22

... p m

1

.

If these requirements are fulfilled, the realization of F(p) is possible. There are various methods of realization. Foster’s method is to represent F(p) as an equation (1.2). The circuits, realizing each term in (1.2), were abov above e considered. The inconv inconvenience enience of the method is the necessity of calculating the roots of the denominator denominator.. In Caur’s method the necessity of calculating the roots of the denomi nator disappears. The essence of this method is gradual separating parts Ap or D/p first from F(p), and then from the residuals after separating the pre 15

vious part, with the subsequent realization of separated parts as an induc tive tiv e coil or a capacitor capacitor.. Let F(p) have a pole p = ∞. It means that the degree of polynomial of the numerator is more than the degree of polynomial of the denominator per unit. Let’s suppose that F(p) = Z(p). Having divided the numerator by the denominator,, we separate an integer part1pA. We get Z(p) = 1Ap + Z1(p). denominator The degree of polynomial in the denominator of3Z (p) is more per unit. Hence, the inv inverse erse functionYp () = 1

1 Zp () 1

has the degree of numerator

1 == Yp () more than the degree of denominator per unit. Hence 1 Zp () 1 +Y Ap 22

() p → Z1() p =

1 +Y Ap 22

()p

.

1 == Further by analogy we getZp () 23 () Yp 2

A p+ Z3 () p .

We continue this procedure until the residual is equal to zero. According to such an operation it is possible to represent Z(p) as a chain fraction: 1 Zp() =+A1 p . 1 + Ap 2 + ... Ap 3 ........................... 1 + Ap k −1 +0 Ap k From here we can see that it is possible to realize F(p) = Z(p) with the help of the circuit shown in Fig. 1.15.

Fig. 1.15. The procedure of synthesis of the circuit corresponding to the chain fraction

16

If k is an even number (k is the exponent of numerator), the circuit will be, as shown in Fig. 1.16, if k is an odd number number,, the circuit will be, as shown in Fig. 1.17.

Fig. 1.16. The case (1) when k is an even number

Fig. 1.17. The case (1) when k is an odd number

If Fp() ==Y ()p

Gp() Qp()

and the degree of polynomial G(p) is more than

the degree of polynomial Q(p) per unit, if we acted similarly similarly,, we should re ceive ceiv e either the circuit shown in Fig. 1.18 for the case, when k is an ev even en number,, or the circuit shown in Fig. 1.19, when k is an odd number number number.. If the degree of numerator (n) is less than the degree of denominator per unit, if we add the termap n+2

n +2

(at an +2 = 0 ) to the numerator numerator,, it is

Fig. 1.18. The case (2) when k is an even number

17

Fig. 1.19. The case (2) when k is an odd number

possible to use the same method formally possible formally.. How However ever,, in this case there have e1L= 0 in the circuits in Figs. 1.16, 1.17 and will be A1 = 0, i.e. we shall hav C1 = 0 in the circuits in Figs. 1.18, 1.19. Let’s consider a case now when F(p) has a pole p = 0. It means that the polinominal of deno denominator minator has an odd degree and the polinominal of nu merator has an ev even en degree, in this case the degree of polynomial of the denominator is less than the degree of polynomial of the numerator per unit. In this case D/p is separated gradually and the chain fraction, got by this wa way y, has a form: D Fp() =+ 1 p

1 D2 p

+

.

1 D3 p

+ ........

Dk −1 p

+

1 Dk p

+0

Fp() == Zp() The example. There is an input function of the circuit

=

p53++68pp 42++ pp 43

. It is necessary to construct a circuit.

Let’s present the function of circuit as Z(p) =1A p + Z1(p), where A = 1 a == n bm 18

1; Zp () = 1

25 p3 + p 42++

pp 43

.

Let’s find the admittance Yp () 11 +Yp() , where Yp() 2 2

31 =+ pY 23

3

25 pp +

3 1 ; Yp() = p+ 2

3 2 p +3 = 2 . 25 p3 + p

Then we calculateZp22()

where

42

pp ++

14 == Zp () 1

12 == Yp () 2

pp Zp33() == ; Y ()p 3 2 p +3 2

2 +5 pp 4 ; Zp () = p+ Z3() p, 3 2 3 p +3 2 13 / 2 2 +3 3 3 = = p+ ; Y () p = Zp p 2 p 3 () 3

calcul ulate ate Z ()p = ()p . We calc

44

() Yp 4

=

p

.

== 1; Thus we hav have e the following parameters of the circuit: LC 12

43 LC == ; 34 32

1 ; 2

1 ; L 5 = . This circuit is shown in Fig. 1.20. 3

C2 = 1/2

Fig. 1.20. Example

1.5. The realization of the input function ha having ving complex roots of the denominator Let Zp() =

2 ++ 52 pp

2

be. The numerator and the denominator hav have e ++ 41pp complex roots. Brune Brune′′s method is used to realize Z(p). According to this method we shall adduce Z(p) to a form of minimal activ active e resistance, i.e. to Re[Z Z(jω0)]. the form of Z(p) –Rmin , where Rmin = Re[ 2

19

For calculating the frequency ω Re[[Z(jω)] = 0, we shall calcu 0, when Re late a real part of Z(jω). The real part of Z(p), when p = jω, i.e. Re[ e[Z Z(jω)], is calculated as fol lows. This part, as a rational fraction, should hav have e terms with ev even en expo nents relativ relatively ely to jω, since only in this case, when p = jω, the function will be real. Therefore we shall represent Z(p) as the sum of rational fractions consisting of terms with ev even en and odd exponents: Z(p) = N(p)+ )+M M(p). As Zp() =

Np () + Mp() 11

Np22() + Mp ()

, when we multiply the numerator and the

denominator by N2(p)–M 2(p), we shall get:

Np() N ()p −−Mp1() M 2()p Zp() =+ 12 22 Np22() −−Mp () where

() N Np 12

()p 22 Np22()

− Mp () M ()p 1 2 − Mp ()

N 2()p Mp1()

Np1()

22 Np22

()

Mp ()

Re Z ()p

==Np()

M 2() p

p= j ω

,

.

According to this formula for the considering case we have:

()52pp ++ ( 4 1) − 2p⋅ p ()41pp + − 22

Re Zj() ω=

42

20pp ++11 = 16pp ++7

24

22

=

2 2

1 pj=

= ω

pj= ω

42 20ω− 11ω+ 2 . 42 ω−ω + 16 7 1

Let’s differentiate the receiv received ed equation and calculate the extremums of the function

()80ω− 22ω ( 16ω −7ω +1) − {} ()16ω−ω7 +1 64ω− 14ω ( 20ω − 11ω + 2) () − =0 ()16ω−ω7 +1 34

d Re Zj () ω= d

2

2

42

34

2

42

20

2

1 ± 1 = R in corresponds to this . Re Zj ω= Hence we find ω= 0 0m 2 frequency.. frequency 22++ 52 pp 2 p + p+ 1 − = −1.= = Z ()p Zp R Thus we hav have e () min 1 22++ 41 pp 41pp++ 1 1 = j ω L where L = −1. At pj= ω0 = j we hav have eZj10 ω= −j 0 0 0 2 2

()

()

Then it is possible to represent1Z (p) as Z1(p) = pL0+Z 2(p) (the correspond ing circuit is shown in Fig. 1.21) whereZp21()

+= p

32 ++ 422 pp 2 ++ 41 pp

=Z

()p − pL0 =

2 ++ pp 1 2 ++

+

41pp

p +1 .

At p = jω0 we hav have eZp2 () =0 .

Yp () 2

== where BY 32

()p

14 == () Zp 2

2 p +

2 ++ pp

()

21+ ω pp ()42 + 2

1 2

1 ; AY= 2 42

p p2 = −

1

1 2

=

Bp 3 1 p + 2

+

2

()p ⋅(4

p+2)

A4 42p+

p= −

1 2

,

= 2.

1 p 2 =+

2 = Y ()p + Y4 () p. p+ 2 + 1 42 p 2 The realization of Y3(p) is carried out by the abov above e mentioned method as a series circuit consisting of3 L= 2 and C3 = 1, and the realization of Y4(p) is carried out as a series circuit consisting of 4 =L 2 and R 4 = 1, since

Therefore Yp () 23

Yp () = 4

1 . 21 p+ 21

The resultant circuit corresponding to the input function

Zp()

=

2 ++ 52 pp 2 ++ 41pp

2 ,

is shown in Fig. 1.22.

Fig. 1.21. The circuit corresponding to 1Z(p) = pL0+Z 2(p)

The negativ negative e inductance 0L can be realized by using the transformer with the coupled factor k = 1.

Fig. 1.22. The circuit corresponding toZp() =

2 ++ 52pp 2 ++

41pp

2

2. ELECTRIC ELECTRICAL AL CIR CIRCUITS CUITS WITH DIS DISTRIBUTED TRIBUTED PARAMETERS (THE STEADY REGIME) 2.1. Electrical circuits with distributed par parameters ameters Earlier we considered the circuits with lumped parameters, and we ad mitted that R, L and C were lumped on certain pieces of the circuit (in resistors, coils and capacitors). In cases, when the time of transmitting electromagnetic waves along the circuit is comparable with that time during the current and voltage change by the value, which is a noticeable part of their total changing in the considering process, the abo above ve indicated admission is prohibited to do. It’s necessary to consider the circuit as a circuit with distributed parame ters. In this case the current and voltage are functions of tw two o independent variables: t (the time) and x (the coordinate). The equations describing processes in these circuits are equations in partial derivatives derivatives.. Examples of circuits with distributed parameters: 1. The pow power er lines, 2. The communication lines, 3. The high frequency cable communication lines, 4. The windings of transformers and electrical machines when a pulse voltage acts. If the parameters of a circuit are distributed along the length uniformly uniformly,, such circuits (lines) are named homogeneous. We introduce the concepts of running parameters (L, C, R, G and M per unit length of the line) for homogeneous lines. In engineering calculations the dependence of parameters on frequen cy is not taken into consideration (L, C, R, C, M = const).

2.2. The equations of a line with distributed parameters Let’s consider a tw two o wire homogeneous line (Fig. 2.1). Here L and R are the inductance and resistance of a pair of wires per unit of the line 23

S

Fig. 2.1. The tw two o wire homogeneous line

length. C and G are the capacitance and leakage conductance betw between een the wires per unit of the line length. The coordinate x is read off from the be ginning of the line. According to the continuity principle of current we shall write the equa tion for the surface S:()−ii +

+

∂∂ iu dx + Gdxu + Cdx ∂∂ xt

=0, here Cdx

∂u ∂t

is the bias current, Gdxu is the conduction current. This equation will be transformed into −

∂∂ iu =Gu + C ∂∂xt

.

(2.1)

The voltage betw between een the wires depends both on t and also on x, since two o wires on each piece of the line there is a voltage drop du 1+du 2 in the tw (dx dx)) (Fig. 2.2). This voltage drop consists of the voltage drop across the

Fig. 2.2. A piece of the line

24

∂i caused by the ∂t ∂i inductance Ldx of the pair of wires, i.e.du12+=du Rdxi + Ldx . ∂t Let’s consider a loop. Due to Kirchhoffs second law we shall write Ldx resistance of the pair of wires and the voltage drop

()−uu+

+

∂∂ ui dx + Rdxi + Ldx ∂∂ xt

=0 or −

∂u ∂i = Ri + L . ∂x ∂t

(2.2)

In the common case, when we have a n wire aerial line located abo above ve the ground surface, it is necessary to take the mutual EMF EMFss and bias cur rents betw between een the considering wire and the neighbouring wires into con sideration in these equations for each of these wires. Then we get 2n, the so called telegraph equations:

−

−

∂∂iu kk ∂∂ xt

∂∂ uikk ∂∂ xt

= mn

= Ri

== mn

=Gu + kk

kk

+L

k

+

∑ M km

∂i m

m=1

()

∑∑Gkm uk − um + Ck

== mm 11

∂t

,

mn

+

() km− u

du Ckm

∂t

,

here k is the number of the wire, all the parameters are calculated taking the ground into consideration.

2.3. Solving equations of an uniform line (the steady sine wave regime) Lets admit that the current and voltage change with a frequency ω. Let’s write the equations of a line using the complex method

−

dU = RI + j ωLI , dx

(2.3)

dI =GU + j ωCU, (2.4) dx since the voltage and current depend only on the coordinate ·x, I(·U= f (x)) let’s write the total deriv derivativ atives es instead of partial derivativ derivatives. es. We differentiate the equation (2.3) by x and use the equation (2.4): −

25

2

dU dx

where γ =+ ()Rj

2

ωL (G + j ωC) U = γ2U,

=+()Rj

ωL (G + j ωC) = α + jβ.

The solution has a form −γγ xx UA=+ 12 ee A

.

(2.5)

From the equation (17.3) we get: IA= −

γ+ 1 dU = Rj+ ωL d x Rj + ωL

()

− γxx ×−AAee 12

where Z =

() ee −A 1 = ()AAee − Z

γ

− γxx 12

γ

− γxx

G j ωC × Rj + ωL

= γ

,

12

Rj+ ωL wave, e, Z is the , γ is the propagation coefficient of the wav Gj+ ωC

wave impedance or the characteristic impedance of the line, α is the attenu ation coefficient (α > 0), β is the phase coefficient (β > 0). Let’s designate the values of current and voltage at the beginning of the line (x = 0) with index «1», and at the end of the line (x = l) with index «2». Let’s find A1 and A2 in the equation (2.5), in order to do it we should consider the beginning of the line. 1 = At x = 0 U·1 = A1+A 2 and IA 11 Z 11 A =+ U 11 22

()

IZ 1

()

− A , then we shall get: 2

and A 2 =

()U 1 − IZ1

.

Hence

11 − γxx + U −I Z UU=+ 11 I Z ee 11 22 11 − γxx 1 − U −I Z IU=+ I Z ee 11 1 1 Z 22

()

()

()

()

γ

, γ

.

or,, taking hyperbolic functions into consideration, or

UU= 26

ch γx − I Z shγx , 11

I = I 1 chγx −

U1 Z

shγx .

(2.6)

We get the values of·2Uand I·2 (the end of the line) if we insert x = l.

ch γl − I 1Z shγl , I

=I 2

chγl − 1

U1

shlγ . Z · and · · · In these equations we shall express 1 U I 1 by U2 and I 2 = UU 21

shγl I 2 chlγ . Z Two last equations are the equations of a tw two o port netw network ork in A param eters. The constants of this tw two o port netw network ork are equal: = UU 12

ch γl + I 2Z shγl , I 12=+ U

shγl B = Z shγl ; C = , and Z AD − BC = ch22γ l −sh γl =1.

chγl ;

== AD

Like any tw two o port netw network, ork, the line can be represented by a Тfigura tive tiv e equivalent circuit or a Pi figurativ e equivalent circuit. It is reasonable to represent the line by a Тfigurativ e equivalent circuit or a Pi figurativ e equivalent circuit if we are interested only in the currents and voltages at the input and at the output of the line. If it is necessary to know the distribution of the current and voltage along the line, we should represent it like a chain circuit. The more we hav have e sections the more the solution is exact (usually we take 10—20 sections).

2.4. Running wa wave ves s Let’s consider the equations (2.6) and introduce the following designa tions:

1 =+ UU ϕϕ 2 1 UU = 2 Ψ

() 11 I Z () 11− I Z

−

−γξ

γxx =U eee 11

γγ xx =U ψψ eee

At γ = α + jβ we hav have e =+ ϕψ U UU

j

=Uϕ

=U

−

e γx ,

j η γx

e .

11

− γxx =Uϕ ee +Uψ 11

γ

ξ−β ξ− β =Uϕ e− γx ejx()

+

1

ax j ()η+βx + U ψ ee . 1

Let’s pass to the originals: 27

uu=+ ϕΨ u = 2e Uϕ

− αxx

sin()ωt +ξ − β x +

11

2UΨ eα

sin()ω t + η +β x .

It is visible that u is equal to the sum of tw two o components: u uψ. ϕ and At x = const uϕ is a sinusoidal function of time. Let α = 0 (e–αx = 1) be, then, if t = const, ϕu (Fig. 2.3) is distributed

λλ = along the line by the sinusoidal law with the wav wave e length

2π . β

Fig. 2.3. Running direct wav waves, es,1 t< t2 < t3

The voltage wav ave e is said to mov move e along the line from the beginning to the end with a constant speed vv =

ω . Since the phase of oscillations β

remains constant, v is named the phase speed speed.. Such wav waves es are named running waves waves.. At α > 0 the presence of the mul tiplier e–γx shows that the peak of the wav wave e attenuates by the exponential law (α is the attenuation coefficient coefficient)) during its mov movement. ement. Since the voltage phase changes with changing x, the coefficient β characterizing this chang ing is named the phase coefficient coefficient.. = uU By analogy it is possible to show that ΨΨ 2e

λ= presents a wav wave e with the length

αx

sin(ωt + η +β x )

1

2π , running along the line with a speed β

ω , i.e. from the end to its beginning. The peak of this wav ave e attenu β ates (eαx) by the exponential law during its mo movvement from the end to the beginning. v=−

28

The wav wave e uϕ is named the direct wave wave.. The wa wave ve u named the return ψ is wave. Similarly we can write for the current·:I 11 U 11+I Z e− γxx ;I Ψ = − U 11I−Z eγ ϕ 22 ZZ or for the instantaneous values of ii ϕ=+ii ψ, where ϕi is the direct wav wave, e,ψi is the return wa wave. ve. We shall calculate: II =+ ϕΨ I

, whereI

()

=

Uϕ

()

==ZZ;

UΨ

− .

IIϕΨ

(2.7)

Physically it is possible to explain the appearance of return wa waves ves by reflection of direct wav waves es from the end of the line. Therefore the direct wav ave e is also named the impinging wave wave,, and the return wav wave e is named the reflect ed wave. Their ratio is named the reflection coefficient coefficient:: А) the reflection coefficient of voltage fro from m the end of the line:

qu =

UΨ

2

Uϕ

,

2

B) the reflection coefficient of current from the end of the line: IΨ 2 qi = . Iϕ 2

Let’s suppose that the line with the wav wave e impedance is connected to the load which has the impedanceload Z . =+ ϕΨ U We hav have eUU 22

;

22

I = I ϕ +I 2

ψ

2

UU ϕψ = 22 − ZZ

at the end of

the line. = From here 2UU Ψϕ 22

=+IZ 2

() load

− I Z =I

22

Uψ Z ; where q == 2 u U ϕ 2

2

()Z load −Z

− ZZ load + ZZ load

;

U 2

=+ U 22 I Z =

.

29

By dividing the first equation by the second equation (2.7), we shall receive receiv e

qi qu

= −1, i.e. qq= − iu

=

ZZ− ZZ+

load

.

load

Let’s consider special cases. = 1) If ZZ load

U Uϕ == II ϕ

→q =0; q =0 (there are no reflected wa wavves), therefore u i

Z.

= ∞→ 2) If Zq load

u

=1 and qi = −1, hence:

= Ψ , i.e. the voltage at the end of the line ·2(U а) UU ) is doubled in ϕ 22 comparison with the direct wav ave, e, b) IIϕΨ = − 22

→ The current at the end of the line ·(2I) is equal to

zero. 3) Zqload = 01→ and II2 = 2. ϕ

u

= − and qi = 1. UU = − Ψϕ 22

· = 0, and U 2

IIΨ = 22

ϕ

2

2.5. Characteristics of a uniform line. Conditions for an undistorting line Z = As you can see that the wave impedance of the line

β= the propagation coefficientγ = α + jR

()

Rj+ ωL and Gj+ ωC

+ j ωL (G + j ωC) depend on

frequency. Therefore the conditions of running current and voltage wav frequency. waves es of various harmonics for various frequencies are various. I.e. the periodic non sinusoidal signal is distorted when it is moving along the line. In order the signal not to be distorted it is necessary the wa wavve impedance Z the attenua ω tion coefficient (α) and the phase speedv = β should not be depended on frequency and the phase coefficient (β) should be proportional to frequency frequency.. 30

RG It can be carried out if the condition is kept: = LC

Rj+ ωL Then we getZ == Gj+ ωC

γ=

()R + j ωL (G + j ωC)

=+ RG

= LC

.

R + jω L L L = and C G + jω C C

()RRLC+j ω(G

)

+j ω = LC

() L

+j ω =

jω LC .

Under these conditions the attenuation coefficient and the phase coef RG and β= LC , and the phase speed ficient are minimum: α= min min

ω 1 is maximum vmax == (it is equal to the speed of a running wav wave e in β LC the dielectric surrounding the wires of the line). 8 m/sec. For the cable For the aerial lines Z ≈ 300 300÷ ÷400 Ω and v ≈ 3·10 8 m/sec. lines Z ≈ ≈ 50 Ω and v < 3·10 v λ= (for 50 Hz) = 6000 km. At The wa wav ve length of the aerial line is f 5000 Hz λ = 60 km and it is possible that some lengths of a wav wave e are placed on the line of communication. RG The following inequality > is usually carried out in lines (since G is LC RG small), therefore to achiev achieve e = we must increase the inductance spe LC cially,, connecting inductance coils in the line or using cables which hav cially have e conductors winded by a thin tape with a large magnetic permeability μ. The signal is not distorted and ev even en not attenuated if the following condi tions: R = 0 and G = 0 are carried out. In order to transfer signals into the load from the line without distorting it is necessary to carry out the next conditions: matched), ), 1) Zload = Z (the load and the line are matched 2) if Zload ≠ Z, it is necessary to connect a matching device, for exam ple, a transformer transformer..

2.6. An uniform line at various working regimes In order the solution should be univ universal ersal (for various impedances of a load) let’s place a datum point at the end of the line and replace: x → l – x 31

(x = 0 is the coordinate of the end of the line; x = l is the coordinate of the beginning of the line) in the earlier deduced equations. Then the equations will be: −γγ lx UA=+ 12ee

x = A eγl e−γγ A3 ex + A4 −eγx ,

−γγ lx IZ = A12 ee

−γx x = − A eγl e−γγ A3 ex − A4 e .

(2.8)

At the end of the line (x = 0) → ·U= U·2 and I· = I·2 . have e the following In order to calculate the constants 3Aand A4 we hav · = A +A and I· Z = A –A , hence we getAU=+1 equations U 2 3 4 2 3 4 32 2

()

I 2Z ;

1 −I Z . 42 2 2 The equations (2.8) will hav have e a form:

AU=

()

11 γ− xx UU=+ 22 I Z ee + U 22−I Z 22 11 γ− xx− 1 IU=+ I Z ee U 2 −I 2Z 22 Z 22 or UU=

()

()

()

()

chγx + I Z shγx , 22

I =I

2

chγx +

γ

, γ

U2 Z

, γx . sh

1. Let’s consider an idling regime (all the values should be written with the index «0»). Zload = ∞ and ·I2 = 0. Then these equations become simpler:

= UU 02

I 0=

U 20

(2.9) shγx . Z The input impedance of the line ( at the beginning) is given giv en by

U Z0 == 10 I 10

chγx; 0

Z . thγl

If we analyze the curves corresponding to the equations (2.9), we can 1 notice that in the lines whose length does not exceedλ , at the idling 4 32

regime the working current decreases and the working voltage increases to the direction from the beginning of the line to its end. 2. Let’s consider a short circuited regime (all the values should be written with the index «sc sc»). »). For this regime the following equations are · shγγx ; right: Zload = 0 and U2 = 0. The equations will hav have e a form:·sc U= I ·2scZsh · · I sc = I 2scchγ x. U1sc The input impedance of the line (at the beginning) is giv given en Z bysc == I 1sc = Zlthγ . Having determined Zsc and Z0 from the experiments of the short cir cuited regime and the idling regime, it is possible to calculate Z and γl

ZZ=

Z and thγl = 0 sc

Z sc Z0

.

We can get any working regime of a line connected with Z by the load superposition of the corresponding regimes: the idling regime and the short circuited regime. In this case the voltage and current are calculated as fol lows:

where thσ=

UU=

chγx + 22

II =

chγx + 22

Z Z load

Z Z load Z load Z

shγx =U

ch ()γx + σ , chσ

shγx

sh()γ+ x σ , shσ

=I

.

2.7. Working regimes of a loss free line It is possible to neglect losses in a line and suppose R = 0 and G = 0 at high frequencies ( ωL >> R and ωC >> G ). Then α = 0, γ = jββ=ω , LC , L and all the equations, deduced earlier earlier,, become simpler simpler.. Z ==z C 1. The idling regime (load Z = ∞ and ·I2 = 0, there is a datum point at the end UU · chγx= U· cosβx, 20 γ = = U of the line). We have ·U = Ix sh j 0 20 20 0 zz

20

sinβx . 33

In this case tw two o unattenuated running wav waves es with equal peaks, mo moving ving in the opposite directions, are superposed. As a result we get the standing waves. In fact cos βx = ±1 at x = 0; λ/2; λ; 3λ/2; and sin βx = 0, therefore in the corresponding points of the line we hav have e antinodal points of voltage (the diagram has minimum or maximum in this point) and nodes of current (the diagram crosses the abscissa axis) (Fig. 2.4). Zload = ∞

Fig. 2.4. Standing wa waves ves of current and voltage in the idling regime

At x = λ/4; 3λ/4; 5λ/4 … we receiv receive e nodes of voltage and antinodal points of current (cos βx = 0; sin βx = ±1). U Zj0 == 10 − zctanβl = In this case the input impedance is determined I 10 = jx , here xl is the corresponding inductiv inductive e reactance. l In this case at 0 < l < λ/4 Z capacitive e character; at λ/4 < λ/2 0Z 0 has a capacitiv has an inductiv inductive e character etc. At l = λ/4, l = 3λ 3λ//4 →0Z= 0 (there is a voltage resonance). At l = λ/2, l = λ →0Z= ∞ (there is a current resonance). · = 0. The = 0 and U 2. The short circuited regime regime.. In this regime loadZ 2 shγγx= equations, describing this regime, look as follows:·scU= I ·2scZsh = jI·2sczs zsiin βx, I·sc= I·2scchγx = I·2sccos βx. From these equations we can see that standing wa waves ves appear again. In contrast to the idling regime there is a 34

node of voltage and an antinodal point of current at the end of the line (Fig. 2.5). Zload = 0

Fig. 2.5. Standing wa waves ves of current and voltage in the short circuited regime

U The input impedance isZjsc == 1sc I 1sc

zt anβl = jx l, here xl is the corre

sponding inductiv inductive e reactance, i.e. at 0 < l < λ/4 the input impedance has an inductiv inductive e character character,, at λ/4 < l < λ/2 the input impedance has a capac itive itiv e character etc. At l = λ /2, l = λ → Zsc = 0 (there is a voltage resonance); at l = λ/4, l = 3λ/4 → Zsc = ∞ (there is a current resonance). At very high frequencies the short circuited line with the length equal to λ/4, is used as an oscillating loop with a small attenuation and a large input impedance therefore, when there are wav waves es with a small length, this line can be used as an isolator to avoid large losses of energy energy.. 3. The regime of a wattless loadload (Z= jxload). We hav have eUU= II =

cosβx + 22

z xload

cosβx − 22

xload zc

sinβx =U

sin ()βx + σ , sinσ

sinβx =I

cos(βx + σ) , osσ 35

here tanσ=

xload

, thus the standing wav waves es also arise in this regime, but there z is neither an antinodal point nor a node at the end of the line (Fig. 2.6).

Fig. 2.6. Standing wa waves ves of current and voltage in the regime of a wattless load

The input impedance is equal

U Zjxl == 1x I 1x

x oad

At l = λ/4 and lZ= λ / 2 →

tan ()β+ l σ = jzt an ()βl + σ = jx . l tan σ

x

=−

z2

()jxload

and Zx = jxload. As it is visible,

tanβ βl there are neither antinodal points nor nodes. They arise load at x = zctan π equivalent alent to the short circuited when σ=± − βl , Zx = ±∞, i.e. the line is equiv 2 line with the length equal to λ/4. At load x = −ztan ztanβ βl, when σ = −βl, Z x=0 and the line is equivalent to the open circuited line with the length equal to l = λ/4. Thus in this regime the line can hav have e both an inductiv inductive e character and a capacitive capacitiv e character character.. 36

Conclusion. The standing wav waves es appear in all the three regimes of a loss free line. In this case the antinodal points of voltage and current and also nodes of voltage and current are displaced fro from m each other by λ/4. All the three cases are characterized by the absence of pow power er consump tion both in the line and in the load. If there is pow power er consumption in the line, we hav have e running wav waves es in stead of standing wav waves es of voltage and current.

3. TRANSIENTS IN CIR CIRCUITS CUITS WITH DIS DISTRIBUTED TRIBUTED PARAMETERS Here are examples from life: the switching on and off of a line, the in fluence of thunder ov overvoltages ervoltages on a line etc. Currents and voltages in co com m munication lines, as a rule, hav have e an acy acyclic clic character character.. The transients in these cases are described by the equations with partial derivatives. derivativ es. The solution can be carried out by both the classical method and the operational method.

3.1. Transients in an uniform undistorting line (the classical method of calculation) The transients in an uniform undistorting line are described by the equa tions −

∂∂ ui = Ri + L ∂∂ xt

−

;

∂i ∂u =+ Gu C . ∂x ∂t

(3.1)

RC = GL since the line is undistorting. Let’s introduce the new values:δ=

Then we hav have e

uu ∂∂ uu ∂∂ == 11e;− δtt ∂∂ xx

RG u = , ui = a nd 11 − δtt LC ee

∂∂t

t

e− δ − δu1 e;− δt

=

i −δ

.

∂ i ∂ i1 − δt = e; ∂x ∂ x

∂i ∂i1 − δtt = ee − δi1 − δ . ∂∂ tt Let’s substitute these equations into the equations (3.1) and divide them by e–δt , we shall get: −

38

∂∂ ui11 = LC ; ∂∂ xt

−

∂i 1 ∂∂xt

=

∂u 1

.

Let’s differentiate the first equation by x, and the second equation by t: ∂∂22ui 11= − LC ; 22 ∂∂ xt ∂∂ xt

−

∂∂22 uu From here we shall get 11= LC ∂∂ xt22

∂ 2i ∂∂xt

1

∂ 2u

=

1

.

, since CL =

1 v

2

.

∂∂22uu 11= 2 v The equation will be transformed into the wave equation equation::22 ∂∂ tx Let’s introduce the new variables ξ = x–vt; η = x+vt taking ∂ξ ∂η == 1; 1; ∂∂ xx into consideration, then we get

∂∂ uu 11=+ ∂∂ ξ xt

∂ u1 ∂η

∂ξ = −vv; ∂t

∂ u1

;

∂

∂∂22 uu 11=+ 22 ∂∂ ξ x

∂η = ∂t

.

,

∂u ∂ u1 = −vv 1 + ; ∂ξ ∂η

∂ 2u ∂ 2u 1 2; 1 + 2 ∂ξ∂η ∂η

∂∂22 ∂ 2u1 ∂2 uu 2 u1 11= 22 − 2. +v vv 22 ∂ξ∂η ∂∂ ξ ∂η 2 t Let’s substitute these equations into the wav wave e equation 2 uu ∂∂

11== 0

∂ξ∂η

or

∂ ∂ξ ∂η

0.

Let’s integrate the deduced equations

∂u 1

= γ()η= and ud ∫ γ()ηη + ϕξ( ) = ψ()η+ ϕξ( ) 1 ∂η and return to the variables x and t and write the following equation: ux = ϕ() −vt + ψ()x +vt . 1

(3.2)

From here we get an equation for the voltage betw between een the wires of the line − δt ux= ϕ() − vt + ψ()x + vt e. 39

Let’s find i1, we should substitute (3.2) into the equation We get i1

CC

xt

t

t

C Lx since

and

xx

x

iu 11 xt

C

.

Cv

,

. C vt L 0. For the current in the line we

Having integrated this equation, we shall findix 1

xv t

f t . Let’s admit that f(t) C L

shall getix

vt

x vt e.

t

RG , named the attenuation

Let’s introduce a new quantity

coefficient of an undistorting line, then the receiv received ed solutions are written as follows: ux

vt ee

xx

x vt

; i

C L

x vt ee xx

x vt

.

The functions and differ from and in the previous equations by vt)) and e (x vt vt).) the multipliers e (x vt The general solution is receiv received. ed. The form of the functions and is defined by the conditions of a concrete problem.

3.2. Transients in an uniform undistorting line (the operational method of calculation) Since the originals areui,

f t , x , the operational images Fp , x are

determined as follows:

ut , x

ii ii

40

U px ,

∫∫ ut , 00

xe

pt

d;t

i t, x I px ,

i ,t xe

pt

d.t

∂u i The time derivativ derivative e has an image: = pU ()p, x − u( 0, x) where ux()0, 0, ∂t i is the distribution of u along the line at t = 0. ∂udi The coordinate derivativ derivative e (x) has an image: = Up() , x . ∂xd x i For the current the derivativ derivatives es are accordingly determined as follows: ∂∂ ii ii d = pp = I,()p x . I,() x − i(0, x); ∂∂ txii dx Taking the abo abov ve recorded equations into consideration, the equations of an uniform line in the operational form are written as follows: − −

dU ()p, x

dx dp I,() x dx

= Rp I,() x + pLp I, () x − Li(0, x);

= GU ()p, x + pCU() ,p x − Cu(0, )x.

The receiv received ed equations are ordinary differential equations since they contain one variable x. Let’s set the boundary conditions: x = 0 and x = l and accept the zero initial conditions: u(0, x) = 0 and i(0, x) = 0. These equations hav have e a form: dU ()p, x

= ()R + pL I (px,

);

dI ()p, x

=()G + pC U ( px , ). dx dx Let’s differentiate the first equation by x and take the second equa −

−

2 dU tion into consideration, we shall get

()p, dx

2

x

= γ2Up() , x , where γ =

=+()Rp L (G +pC) . − xx We search the solution for voltage in form: Up() , x =+A12 e γγ A e , = ()p and AF = ()p , and search the operational current in form: where AF 11 22

Ip() , x =

1 Zp()

()A e

− γxx −A 12

γ

e

, where Zp() =

Rp+ L is the operational Gp+ C 41

γ= wav ave e (characteristic) impedance of the line,

(Rp+ L)(G + pC)

is the

operational distribution coefficient. RG The giv given en equations become simpler for the undistorting line = LC

:

Lp , γ = RG + pL C = α + . Cv In this case the operational voltage and current are equal: Zp() =

Up() , x =+ F12(

xx − pp p e vv

)

− αxx

e

F () p e

α

e

,

xx

− pp CC − α xx − Ip() , x = F12()p e vv e LL

F ()p e

α

e

.

Now using Riemanna Mellina`s formula let’s pass to the originals. − αxx + ψ( x + vt) eα , ut(), x = ϕ ( x − vt) e

it(), x =

C ϕ ()x − vt e−αxx− ψ()x + vt eα L

.

Here the term ϕ(x − vt vt)) represents direct wav waves es of voltage and current running with a speed v from the beginning of the line to its end; the term ψ(x +vt vt)) represents return wav aves es running with a speed v fro from m the end of the line to its beginning. In view of the abov above e said the voltage and current through the line can be represented as the sum of direct and return wav waves. es. In any point of the line the ratio of voltage and current for a direct wav wave e is L L , and for a return wav wave e is equal to — . C C The multiplier e−αx (eαx ) speaks about attenuating wav waves es while they are running. The conv conversing ersing energy of electrical and magnetic fields into heat is the reason of attenuating wav waves. es.

equal to

42

3.3. Refraction and reflection of waves at the position of joint of tw two o uniform lines Let the wa wave ve ϕ1 reach the position of joint of tw two o uniform lines, charac terized by wav wave e impedances:1 Zand Z2. Further all the quantities, con cerning to the first line, will be marked by index «1», all the quantities of the second line by index «2». Let’s suppose that before the wav wave e approach es, there is no voltage across the second line at all. The wav wave e ϕ1 passes through the position of joint and begins to mov move e along the second line, not changing its direction, this wa wave ve is named the refracted wave and denoted by2. ϕIf Z 1 ≠ Z2 (the lines are not matched), the part of the wav wave e 1ϕwill be reflected from the position of joint and begin to mov move e along the first line in the return direction, this wav ave e is named the reflected wave and denoted by1. ψ wave e does not appear appear.. At the position of joint of If Z 1 = Z2, the reflected wav the lines the direct, refracted and reflected wav waves es are connected with each other by the following equations: =+ uu u 12 ϕΨ

= uϕ = u ,

11

i1 =

2

− uu ϕΨ 11

ZZ 12

=

uϕ

2

= i2 .

Hence we get: 22 ZZ 22 == uu ϕϕ ZZ ++ 21 21

; uΨ

1

−−Z 1 ZZ21

uϕϕ; ii = 1

2

Z1 Z 1+Z 2

; iΨ = ϕ 1

1

Z1 Z 2 Z 1+ Z

iϕ .

2

1

From these formulas it follows that the refracted and reflected wav aves es

α= have hav e the same form like direct wav waves. es. The values ui

uiϕϕ

22

uiϕϕ

,α =

are

11

uiψψ ; are named named the refraction coefficients, coefficients, the values11==qq ui uiϕϕ 11 the reflection coefficients. From the equations we can see that the refracted and direct wav aves es coin and cide by a sign. The sign of a reflected wav wave e depends on the ratio1 of Z Z2, for example, at Z2 > Z1 the wav wave e of voltage is reflected without chang ing the sign and the wav wave e of current is reflected with changing the sign. The 43

direct wa wav ve of current in the first line decreases and the wa wavve of voltage increases, but no more than twice (Fig. 3.1).

Fig. 3.1. Direct and refracted wa wav ves of current and voltage at 2 >ZZ1

the refracted wav wave e of voltage cannot Even at very large magnitudes of2 Z Even exceed a direct wav wave e more than twice. wave e If Z 1 > Z2, the refracted wave of voltage is less than the direct wav and the refracted wav wave e of current is more than the direct wav wave. e. The wa wavve of voltage changes a sign and the wave of current doesn’ doesn’tt change a sign (Fig. 3.2). As a result the current increases but no more than twice.

Fig. 3.2. Direct and refracted wa wav ves of current and voltage at 2 ZZ1

It is possible to write1i = i2; u1 = R0i 2+u 2 for the position of joint. For the wav ave e to pass through the position of joint it is possible to write − uu ϕΨ 11

ZZ 12

uϕ =+ 2 ;

uu ϕ

11

Ψ

=

R0 Z2

uϕϕ + u . 2

2

Hence, 2Z 2 == uu ϕϕ ZZ ++ R0 21 21

;

uΨ

− ZZ 21 1

+R

0

ZZ21++

R0

uϕ . 1

the refracted wav wave e of volt We can see that the presence of0 decreases R wave euψ age and it can be decreased ev even en to «0». 2At> ZZ1 the reflected wav

1

uϕ . increases when 0Rincreases, but it can’t exceed 1

51

== i22 The pow power er emitting in R0, is pR 02

The pow power er of the direct wav wave e ispϕ = 1

R0

u .

ϕ Z22 2

uϕ2

1

Z1

.

A considerable part of the pow power er of the direct wa wavve can be absorbed by wavve is small in the resistance R 0 only at Z1 >> Z2, i.e. when the refracted wa comparison with the direct wav ave. e. In order the ov overv ervoltage, oltage, arisen on a section of the line, shouldn’t movve along the whole length, we connect separate sections of the line with mo the help of special circuits consisting of activ active e resistances (500—600 Ω) and inductiv inductive e coils. These activ active e resistances and inductiv inductive e coils are con nected in parallel. These inductiv inductive e coils hav have e a small resistance for the cur rent of commercial frequency frequency,, but they hav have e a large resistance at the first moment when the wav ave e comes. = If we take R0, equal to Z1+Z 2, we getuu 21 ϕϕ

Z2 + ZZ 12

. In this case the

refracted wav wave e of voltage is less twice than when there is no resistance R 0.

4. NONLINEAR CIR CIRCUITS CUITS OF A DIRECT CURRENT Electrical circuits are named nonlinear if all the parameters or even one of them depend on the magnitude of current and voltage. The processes in such circuits, as well as in linear circuits, are described by the equations based on Kirchhoffs equations. How However ever,, the equations are nonlinear and the methods of their calculation hav have e their specificity specificity.. In a case of a direct current the equations of a nonlinear circuit will be nonlinear algebraic equations. In order to calculate them we should have characteristics of all the nonlinear elements, i.e. the dependences of volt age across the terminals of each element on the current, running through it. These dependences are named the voltage current characteristics (VCC) and they are usually written as follows u = f(i). Sometimes we use depen dences of current on voltage i = f(u), named the current voltage character istics.

4.1. Nonlinear elements and their characteristics At the action of constant EMF EMFss and voltages in the circuit the magnitude of a direct current in it depends on the resistance R and conductance G of the elements of the circuit. These parameters are basic. As to the capacitance and the inductance in case of nonlinear circuits of a direct current, they pla playy a role only at solving the problem of stability of the regime. The resistance and conductance of many nonlinear elements also hav have e a basic meaning in circuits of an alternating current. In this connec tion let’ let’ss consider such nonlinear elements and their characteristics which basic parameters are resistance and conductance. For the element, characterized by a constant resistance, the VCC is a direct line (Fig. 4.1). u R == tan α = const. i The characteristics of nonlinear elements areFig. 4.1. The VCC of linear resistance subdivided into static and dynamic. The char 53

acteristics, got at a very slow (indefinitely slow) changing of current or volt age, are named the static characteristics characteristics.. The dynamic characteristics show the connection betw between een voltage and current when they change fast. These characteristics can differ from the static characteristics, for example, be cause of the thermal inertia. There are concepts of static and dynamic resistances and also static and dynamic conductances. The static resistancestRat the giv given en current is the ratio of voltage corre sponding to this current by the static characteristic, to the magnitude of this current (Fig. 4.2). u == Rk st i

tan α;

v where k = , v, a are the scales of voltage and current. a The value, inv inversed ersed to the static resis tance, is named the static conductance.

Gst =

1 . Rst

The dynamic resistancedRat the giv en point of the dynamic characteristic is the deriv derivativ ative e of voltage by the current in the giv given en point of the dynamic charac teristic. The quantity quantity,, inv inversed ersed to the dy Fig. 4.2. The ascending VCC of namic resistance, is named the dynamic nonlinear resistance conductance dG. Let the dynamic characteristic coin cide with the static characteristic. Then the dynamic resistance can be found by the giv given en static characteristic as follows:

Δud u = tanβ, i Δ→ i 0 Δid where β is the angle between the tangent to the dynamic characteristic and the abscissa axis. ==lim Rk d

Gd = 54

1 . Rd

All the mentioned parameters st R, Rd change from a point to a point, i.e. they de pend on the current magnitude. For the pas sive siv e elements, i.e. which dont contain the sources of energy energy,, alwa always ysstR> 0, Gst > 0, but Rd, Gd are positiv positive e values only for the points, lying on the ascending part of the characteris Fig. 4.3. The declining VCC tic and they are negativ negative e values for the points of nonlinear resistance of the declining part (Fig. 4.3).

4.2. Current v oltage characteristics of some nonlinear elements 1. The junction diode. Its current voltage characteristic is giv given en in Fig. 4.4. When a source of a direct current (Fig. 4.5) with an EMF greater than the barrier potential drop, is joined with its positiv positive e terminal to the p semi conductor (P) and its negativ negative e terminal to the n semiconductor (N) posi tive tiv e charges (holes) are urged across the p n junction from P to N and negative negativ e charges (electrons) from N to P.

Fig. 4.4. The junction diode charac teristic and symbol

Fig. 4.5. The junction diode (forward bias)

We can understand the mo movvement if we consider the positiv positive e terminal of the source to repel holes in the p semiconductor and the negativ negative e terminal to repel electrons in the n semiconductor . Thus the appreciable current is ob tained. Now the p n junction is said to be forward biased . When the applied voltage is increased, the current increases along the curv curve e OA (Fig. 4.4). When the terminals of the source are rev reversed, ersed, only a very small current runs (Fig. 4.6). In this case the p n junction is said to be reverse biased . This 55

time only the minority of carriers: electrons in the p semiconductor and holes in the n semiconduc tor are urged across the p n junction by the source. Since the minority of carriers are ther mally generated, the magnitude of the rev reverse erse current (the segment OB) depends only on the temperature of the semiconductors. It may be also noted that the rev reverse erse bias potential drop increas es the width of the depletion layer (the narrow re gion or la layyer at the p n junction, which contains Fig. 4.6. The junction electrons and holes), since it urges more electrons diode (rev reverse erse bias) in the p semiconductor and holes in the n semi conductor are further awa awayy from the junction. The voltage current characteristic (the curv curve e AOB) in Fig. 4.4 shows that the p n junction acts as a rectifier rectifier.. It has a low resistance for one direc tion of the potential drop and a high resistance in the opposite potential drop direction. It is called the junction diode diode.. In the diode symbol in Fig. 4.4, the low resistance is from left to right (tow towards ards the triangle point) and the high resistance is in the opposite direction. The junction diode has sev several eral ad vantages; for example, it needs only a low voltage source to work; it does not need time to warm up; it is not bulky; and it is cheap to manufacture in large numbers. 2. In high voltage engineering tyrite nonlinear elements, made of a ce ramic material (the tyrite), are used. The characteristic of tyrite is shown in Fig. 4.7. The tyrite resistance decreases, when the voltage increases, i.e. the con ductance increases. Such dependence of conductance on voltage allows to use tyrite elements to protect high voltage devices of electrical pow power er sta tions, transformers of substations and etc. against ov overvoltages. ervoltages. We place the so called tyrite discharger (Т) (Fig. 4.8), connected to a spark gap in series, parallel to the protected device (N), betw between een the line of an alternating current of high voltage (HV HV)) and the ground. At the nominal voltage the spark gap is not broken down and the current does not run discharger.. When the voltage across Fig. 4.7. The VCC of tyrite through the discharger 56

the spark gap exceeds the nominal value, the spark gap is broken down and a large cur rent runs through the tyrite discharger discharger,, as its resistance drops sharply sharply,, when the voltage increases. As a result the line (HV HV)) is dis charged into the tyrite discharger (Т) and the voltage drops across the line. Thus the dis charger resistance increases and the current, running through it, decreases. The sharpFig. 4.8. The use of the tyrite decreasing of current leads to the interrup nonlinear element to protect high voltage devices tion of a discharge in the spark gap and, hence, it leads to the interruption of the cur rent in the discharger circuit. 3. The electrical arc, which is a nonlinear element of electrical circuits, has a considerable importance in practice. In Fig. 4.9 the electrical arc (EA EA), ), burning between the coal electrodes under the atmosphere pressure in the air air,, is schematically shown. An activ active e part of the cathode (C), emit ting electrons, has a temperature equal to ~ 3000 °С. A part of the anode (A), bo bombard mbard ed by electrons, has a temperature equal to ~ 4000 °С. The arc itself places between the active activ e parts of C and A, its temperature is equal to ~ 5000 °С. In the arc zone the gas is in the ionized state, the main carriers of cur Fig. 4.9. The electrical arc rent are electrons. Nowada Now adays ys the electrical arc is used as a light source in projectors and projective projectiv e devices. In metallurgy pow powerful erful arcs are used in arc furnaces. The electric welding by an electrical arch is spread widely widely.. The electrical arc has a nonlinear characteristic shown in Fig. 4.10. We can see that if the current increases the voltage of the arc drops.

Fig. 4.10. The VCC of an electrical arc

57

4.3. The calculation of simple circuits with passive nonlinear elements The graphic method of calculation а) The series connection of nonlinear elements. The circuit, which is necessary to calculate, is giv given en in Fig. 4.11. Two nonlinear elements are connected in series in this circuit.

Fig. 4.11. The series connection of nonlinear elements

The characteristics of the nonlinear elements1 = u F1(i 1), u2 = F2(i 2) are given giv en as plots in Fig. 4.12. In this case by Kirchhoff`s laws it is possible to write: u = u1+u 2, i 1 = i2 = i. So summing up the ordinates of the characteristics1u= F1(i 1) and u2 = F2(i 2), we build a characteristic u = F(i). Hav ing this characteristic, it is easy to find the cur rent i and voltages 1u, u2 for any regime. The ex ample for the voltage u =*uis shown in Fig. 4.12. This method can be used for a case of any number of nonlinear and linear elements con Fig. 4.12. The VCCs of nected in series. nonlinear elements b) The parallel connection of nonlinear ele ments (Fig. 4.13). The characteristics of the nonlinear elements u1 = F1(i 1), u2 = F2(i 2) are shown in Fig. 4.14. In this case according to Kirchhoff’s laws we have hav e i = 1i +i 2 , u = u1 = u2 . So summing up the abscises of the curv curves es Fig. 4.13. The parallel u1 = F1(i 1) and u2 = F2(i 2), we build a charac connection of nonlinear teristic u = F(i). elements 58

Fig. 4.14. The VCCs of nonlin ear elements

Fig. 4.15. The mixed connection of nonlinear elements

c) The mixed connection of nonlinear and linear elements (Fig. 4.15). Let the characteristics of the nonlinear elements be known1 (u F1(i 1), u2 F2(i 2)) (Fig. 4.16). The voltage current characteristic of a linear resis tance is written as follows: 3u R3i 3. According to Kirchhoff Kirchhoff’’s laws we have the equations: u 23 u u2 u3, i 1 2i i, u23 u1 u. First we sum up the ordinates of the curv curves es2 u R2(i 2) u3 R3i 3, we curves es build a curv curve e u23 F23(i 2). Then summing up the abscises of the curv u1 F1(i 1) and u F curve e u F(i). Having the given curv curves, es, 23(i 2), we build a curv it is possible to find all the voltages and currents if one of these voltages or one of these currents is known. d) The calculation of simple nonlinear circuits, containing EMF sources (Fig. 4.17). The characteristic uab F(i) and the magnitude and direction of the EMF (e) are giv given. en. By Kirchhoff Kirchhoff’’s second law law,, taking the direction of a path tracing into consideration, we hav have: e: ac u uab ubc uab ebc.

Fig. 4.16. The VCCs of nonlinear elements

Fig. 4.17. The simple nonlinear circuit, containing an EMF source

59

have e a case shown in Fig. 4.18. Let ebc > 0 be. Then we hav The Fig. 4.19 is corresponding to the case when e 0. bc < We can see that the presence of the EMF source, connected with the nonlinear element in series, leads to the appropriate parallel displacing of the characteristic of this nonlinear element. Therefore the calculation of nonlinear circuits, containing EMF sources, is made by the same methods as the calculation of passiv passive e nonlinear circuits.

Fig. 4.18. The VCC of a nonlinear circuit (ebc > 0)

Fig. 4.20. Example

Fig. 4.19. The VCC of a nonlinear circuit (ebc < 0)

The example: It is necessary to calculate a circuit shown in Fig. 4.20. Let ubc = Fbc(i 1), udc = Fdc(i 1) be and the magnitudes and the directions of the EMF given. en. (e1 > 0, e2 > 0) (Fig. 4.21, 4.22) are giv Let’s set up the directions of currents through all the branches. We build a resultant characteristic of all the branches by the abo abovve mentioned method.

Fig. 4.21. The VCC of a nonlinear branch Fig. 4.22. The VCC of a nonlinear branch «ac ac» » (uac = F(i1)) «ac ac» » (uac = F(i2))

60

uuu =+ ac

ab

uu =+ ac

ad

bc

= −e +u , 1 bc

ue = dc

2

+u . dc

(Fig. 4.21) (Fig. 4.22)

We sum up the abscises of the curv curves es ac =uF( i 1), uac = F( i 2), we build a curve curv e u= F(i) (Fig. 4.23).

Fig. 4.23. The VCC of a nonlinear circuit (u = F(u))

4.4. The calculation of simple nonlinear circuits of a direct current by the iter iterative ative method The term «iteration» is originated from a Latin word and it means «rep etition». The iterativ iterative e method of calculating nonlinear algebraic equations is of ten used for calculating circuits with nonlinear elements. To understand the essence of the method we shall consider an equivalent circuit, in which the EMF source Е and resistanceintr (Fig. 4.24) represent any linear part of the initial circuit, i.e. represent some equiv equivalent alent source. Let the external characteristic of the equiv equivalent alent source (U = Eint−I )r coincide with straight line 1 (Fig. 4.25) and the characteristic of the non linear element (U = r (I )·I ) is given by curv curve e 2. If the calculation is made by the geometrical method, the intersec tion of these characteristics (point «а») defines the regime of the cir cuit, i.e. the voltage and current in this regime. If this problem is solv solved ed by the numerical method, for example by the iterativ iterative e method, it is necessary to do the following. Fig. 4.24. The equivalent circuit 61

Fig. 4.25. The characteristics of the source and the nonlinear element

1. We do the so called zero approximation. For this purpose we set up curve e 2. the voltage U I curv 0, equal, for example, to Е and we find the current 0 by Each nonlinear element is linearized at zero approximation. 2. By the equation we calculate: UE= − r I = F () U . (4.1) int We find the impro improv ved value of voltage 1U(i.e. we do the first approxi mation). 3. We find a new value of1Iby characteristic 2 etc. Due to Fig. 4.25 it follows that the iterativ iterative e process conv converges erges to the desired values of current and voltage at the point «а». In the course of math ematics it is prov proved ed that the condition of conv convergence ergence is to carry out the dF ()U < 1 about the desired regime (point «a»). The less is the inequality: dU dF ()U value of the quicker the process conv converges. erges. dU r dF ()U dI = −r = − int , we have the following condition of con int dU dU rd vergence As

−

62

rint rd

S* and according to (7.13) to > 0. As value UFB av FB av E a result the peak of oscillations will increase and approach to the steady value. At increasing the voltage of FB, i.e. at the deviation ofFBUto the * the av * , i.e. right from UFB average erage slope obeys the inequality UFB) < SFB av(S the attenuation coefficient E will become negativ negative e and the valueFB U will decrease, approaching nearer to the steady value again. Thus the point A corresponds to the steady regime. The point B in Fig. 7.6, b corresponds to the unstable steady regime, as *U in the the deviation of the effectiv effective e value FB U from the steady value FB * and direction of decreasing leads toavS(UFB) < Sav E > 0, i.e. to the fur ther decreasing of the voltageFB U. The deviation of the effectiv effective e valueFB U * from the steady value FB U in the direction of increasing will cause its fur ther increasing. The circuit will pass into the following steady regime, marked by point C. The steady condition in the point C is stable. You can notice that the following statement is right: the intersection of * with the curv gives es a stable the direct line Sav curve e of the av average erage slope (UFB) giv avS * steady value U FB, if

dSav U FB dUFB

dS U 0 on this segment and if av FB dUFB

0,

this intersection corresponds to the unstable value. Therefore it is possible to consider the condition

dSav U FB dUFB

0 as the criterion of stability of the

steady regime. Let’s consider the regime of self excitation. We shall change the factor of mutual induction M and watch the process of arising oscillations. This process also depends on choosing a working point on the VCC (the volt age of displacement 0U). The plot of the av average erage slopeavS(UFB), shown in Fig. 7.7, a, corresponds to the choice of the working point in the area of the greatest slope (the voltage of displacement U Fig. 7.7, b). 0 in In case of changing the factor M the magnitude of the av average erage slope LG *S Sav changes. Some straight linesav , appropriated to various M, are M active e oscillator oscillator,, shown in Fig. 7.7, a. At M M 1 oscillations can’t arise in the activ * as Sav > Sav(UFB) and the attenuation coefficient of the loop is positive 1 ( E > 0). It means that any random fluctuations of the voltage FBuwill damp quickly.. quickly 108

* and Increasing M till the value M2 leads to Sav(UFB) < Sav 0. The E 2 * further increasing M decreases the value at the same time the factor E av S becomes negativ negative, e, i.e. E < 0. Thus starting with M2 M undamping oscil * Uarise in the activ active e lations with the corresponding steady magnitudes ofFB * oscillator.. With increasing M the steady effectiv oscillator effective e valueFBUsmoothly in creases. Decreasing M will cause a smooth decreasing of steady magni * . The plot of the dependence of the steady effectiv * U tudes UFB effective e value FB of of oscillations, generated in the activ active e oscillator on M, is shown in Fig. 7.7, b. Such a regime of self excitation of the activ active e oscillator oscillator,, at which the peak of oscillations smoothly increases with increasing M, is named the soft regime of self excitation .

Fig. 7.7. The dependence of an av average erage slope on voltage of the feedback (regime of * self excitation) — a, the dependence of the steady value U M FB on

If we choose the working point on the bending down of the VCC at * (US ) (as it is shown in Fig. 7.4), the plot of the av average erage slope UU FB 00 will be as it is shown in Fig. 7.8, a. At M, equal to M1, M 2 and M3, the presence of small fluctuations of voltage uFB will not lead to setting up a steady peak, as the attenuation *S, equal to S* , coefficient of the loop E will be positiv positive e at the valuesav av1 * and S* . Sav av3 2 * and Only starting with M M4, when Sav(U FB) Sav 0, small E 4

fluctuations of the peak of FB u begin to increase quickly until the steady * sets up. value UFB 4

The further increasing M leads to smoothly increasing the steady peak. At smoothly decreasing the feedback (the factor M) the steady value * will also smoothly decrease. Oscillations will be broken at the value UFB 109

Fig. 7.8. The dependence of: *alue a — an av average erage slope on voltage of the feedback; b — the steady vFB on UM

M 2 (M M2), which is smaller than M4 when the condition of stability * is interrupted. In Fig. 7.8, b the plot of changing the effec Sav(UFB) Sav tive tiv e value UFB, depending on M, is shown. Such a regime, when oscilla tions are excited at the greater magnitude of M and broken at the smaller magnitude of M, is named the rigid regime of self excitation . The advantage of the soft regime of self excitation is the smooth changing of the effectiv effective e * value UFB at changing the factor M. The adv advantage antage of a rigid regime is the high efficiency at the expense of work when there is a cutoff of a collector current. It is possible to unite the advantages of the soft and rigid regimes of self CB into excitation if we introduce the circuit of automatic displacementBR an activ active e oscillator (Fig. 7.9, a) a).. The initial displacement0Uis chosen so that the working point is on the segment with the greatest slope of the volt current characteristic. It corresponds to the soft regime. At increasing the peak of oscillations of FB u in the base circuit at the expense of the nonlin earity of VCC (iB F(uBE)) detecting oscillations will take place. The in creasing of the constant component of the base current B0, Icreating the voltage IB0RB across the activ active e resistanceB,Rwill lead to decreasing the resultant voltage of the displacement 0(– UI B0RB) and, as a result, to the shifting sh ifting of the working point (Fig. 7.9, b) left to the bending down of VCC (ik F(uBE )). The transient is finished (at the appropriate value of RB) by the setting up of the rigid steady regime with a higher efficiency efficiency.. 110

Fig. 7.9. The activ active e oscillator — a; the characteristics of an activ active e oscillator with a circuit of automatic displacement — b

7.2.3. Active oscillators with an internal feedback Earlier one of the forms of the differential equation of an activ active e oscilla 2

du du 2 tor with an external FB (7.9) was receiv received: ed: kk 20 Ek u , 0 dt dt 2 1 GG , GE is the conductance inserted into the oscillato here EE 2C ry loop at the expense of the acting of the external FB. The equality E G G corresponds to the steady regime. The condition of arising oscillations is carried out at GE < 0 and |GE| > G. The comparison of the giv given en differential equation with the differential equation of a single oscillatory loop (7.2) allows to construct an equivalent circuit of an activ active e oscillator oscillator.. It is shown in Fig. 7.3 and differs from the circuit of an usual loop by the presence of negativ negative e conductance in it. The negativ negative e conductance can be receiv received ed not only at the expense of the acting of the external FB, but also with the help of 125nonlinear elements with the VCC, having a falling segment. Electronic devices, which are re sistive sistiv e nonlinear elements with falling segments of the VCC i F(u), are named the devices with negativ negative e resistance. In particular particular,, such a device is a tunnel diode. 111

Activ e oscillators, constructed Active with the help of devices with nega tive tiv e resistance, don’t contain a cir cuit of the external FB and there fore they are named the active os cillators with an internal FB. Fig. 7.10. A tunnel diode: the VCC — a; the equivalent circuit — b The VCC of a tunnel diode is shown in Fig. 7.10, a. The differ di 0 on the segment ab ential conductance is equal:Gu ab.. The equiv du alent circuit of a tunnel diode, corresponding to the falling segment of the characteristic, represents the parallel connection of the nonlinear negativ negative e conductance of the diode G(u), depending on the applied voltage u and the capacitance C active e D of the p n junction (Fig. 7.10, b). The circuit of an activ oscillator with an internal FB, made with the help of an tunnel diode, is shown in Fig. 7.11, a. With the help of the voltage of the displacement 0 U the working point is set up approximately in the center of the falling seg ment of the diode VCC. The blocking capacitanceBLC protects the source of constant voltage fro from m the passing an alternating current with the gener ated frequency frequency.. Having replaced the tunnel diode by the parallel connection of the neg ative ativ e conductance G(u) and the capacitanceD,Clet’s pass to the equiva lent circuit of an activ active e oscillator of an alternating current (Fig. 7.11, b). The capacitance of the equivalent circuit is equal: C k +CCD. The activ active e oscillator oscillator,, constructed with the help of a tunnel diode, is the active activ e oscillator of nearly harmonic oscillations, and it is possible to ana lyze its work by the same wa wayy as to analyze the work of an activ active e oscillator with an external FB. Let’s introduce the concept of av average erage conductance

Fig. 7.11. The active active oscillator with the internal feedback — a; the equivalent circuit of an activ active e oscillator of an alternating current — b

112

effective e of a nonlinear element G av(U1) < 0 by the first harmonic with the effectiv value U1. The plots of the dependences of av |G(U1)| on U1 at various voltag given en in Fig. 7.12. In it the various magni es of the displacement 0Uare giv tudes of conductance of the loop G are shown. The oscillations are excited at |av G(U1)| > G. The steady values*1Uare set up at |G G. av(U1)| Analyzing the dependences in Fig. 7.12, it is possible to hav have e the soft and rigid regimes of self excitation (Fig. 7.13) in activ active e oscillators with an

Fig. 7.12. The dependences ofav |G (U1)| on U1

internal FB. The soft regime of self ex citation occurs at the voltage of dis placement: 0,15V 0U 0,3 0,3V V; the rigid regime occurs at U 0,3V 0,3 V. 0 The steady points A, A and A on the curv curve e of the av average erage conductance |Gav(U 1)|, receiv received ed at U 0,4V 0,4 V, are 0 shown in Fig. 7.12. Oscillations arise at the value of conductance of the loop equal to G (point A ). The effectiv effective e val * ue of oscillations is equal to 1U. In case of increasing the conductance G the Fig. 7.13. The dependences, corre * decreases; the oscilla steady value U sponding to the rigid and soft regime 1 tion stop occurs at G (point A ). of excitation 113

Therefore the block curv curve e in Fig. 7.13 corresponds to the rigid regime of self excitation. The dotted curv curve, e, receiv received ed for the av average erage conductance at U0 0,3 0,3V V, corresponds to the soft regime of excitation.

7.3. The relaxation oscillations The circuit, shown in Fig. 7.14, can be considered as an example of the circuit, in which the instability of the balance state is caused by the pres ence of activ active e resistance, having a fall ing segment of the characteristic, in the circuit. The activ active e resistance R and the capacitance C, shunted by a neon lamp NL , are connected to the source of EMF in series in this circuit. The voltage current characteristic Fig. 7.14. The circuit, generating (VCC) of a neon lamp is shown by curv curve e relaxation oscillations 2 in Fig. 7.15 and the VCC of the resis tance R is shown by straight line 1. They are intersected in the point a, placed on the falling segment of the VCC of the neon lamp. In the circuit there are autooscillations. The relaxation oscillations are the autooscilla tions, arising in the circuit with one energy store, for example in the circuit with one capacitance or in the circuit with one inductance.

Fig. 7.15. The VCCs of a neon lamp and resistance

Fig. 7.16. Relaxation oscillations

The form of relaxation oscillations differs essentially from the sine wav ave e os cillations. In Fig. 7.16 the plot of changing voltage across the capacitance C u as a function of time (the sawtoothed curv curve) e) is shown. The process occurs qualitatively qualitativ ely as follows. In case of switching on the circuit, shown in Fig. 7.14, the capacitance begins to charge. As soon as it charges up to the voltage 1, the U neon lamp is turned on instantly instantly,, and the capacitance discharges quickly through the lamp. When the voltageCureaches the value 2Uthe neon lamp is turned off and the capacitance begins charging again. Further the process repeats. 114

8. CALCULA CALCULATIO TION N OF TRANSIENTS IN NONLINEAR CIR CIRCUITS CUITS Transients in nonlinear electrical circuits are described by nonlinear differential equations. There are no regular methods to calculate these equa tions. If we have analytical calculation in view view,, as a rule, each nonlinear differential equation requires a special method, how howev ever er,, there are approx imate methods, having a more or less common character character.. The graphic method and approximate analytical methods (the method of piecewise linear approximation and the step by step method) are related to them. Let’s consider them. Nowadays Now adays co computer mputer engineering finds a wide application, its use al lows considerably to increase the efficiency of the below considered calcu lation methods.

8.1. The graphic method of calculating tr transients ansients Let’s consider the given method using the example of calculating the switching on the coil with a ferromagnetic core to the constant voltage0)(U (Fig. 8.1). The differential equation, which describes the process of switching on, looks as follows: d ri U 0, (8.1) dt where r is the activ active e resistance of the coil winding. The dependence f(i) is nonlinear be cause there is a magnetic saturation of the core (Fig. 8.2). Let’s admit that the core was demagne tized before the switching on. In this case the dependence f(i) will be characterized by the initial magnetization curv curve. e. It is neces sary to solv solve e the equation (8.1) at f(i),Fig. 8.1. Switching on the coil with a ferromagnetic core shown in Fig. 8.2. 115

Let’s write the equation (8.1) as follows (we separate the variables):

d ∫∫Ur 00 0

t

i

dt t .

Using the dependence Fig. 8.2. The Weber Ampere characteristic

1 curve curv e Ur0 i

f

(8.2)

f( i), we draw a

. For it setting the sequence

1 for each of magnitudes of i, we calculate Ur0 i

, according to this magnitude of current, 1 f f(i) (Fig. 8.2). As a result we get a curv curve e Ur0 i

of them and determine by the curv curve e

(Fig. 8.3). Then the time, corresponding to the magnitude of* , is equal to the crosshatchaded region in Fig. 8.2, according to the formula (8.2). As a result we get a curv curve e f(t) (curv curve e 1 in Fig. 8.4). Now we find a current, appropriate to each magnitude of flux linkage ( ), by this curv curve. e. Then using the curv curve e f(i), it is possible to draw the dependence of cur rent on time (i f(t)) (curv curve e 2 in Fig. 8.4). In Fig. 8.4 the curv curves es f(t) (curv curve e 3) and i f(t) (curv curve e 4) are shown if the dependence betw between een and i was linear and coincide with the initial part of the curv curve e f(i). It is visible that increases much faster at mag netic saturation than in the linear case. In the linear case the equation (8.1) looks as follows:

Fig. 8.3. The determination of time

116

Fig. 8.4. The dependence

f(t)

rr

tt UU L di 00 1e LL ; Lr i U0 i Li 1e dt r r where L is calculated by the linear part of the characteristic

8.2. The method of step by step

, f (i ).

intervals interv als (Eiler’s method)

The step by step method represents the method of numerical integra tion of the first order differential equation. The time interval «S», when the transient is calculated, is divided into N equal parts. The distance betw between een the neighbouring points h k – ttk–1 N is named the step of a computational grid and the points kh t are k named the mesh points of this grid. The derivativ derivatives es are replaced by the fi nite differences and the calculation is solv solved ed in the same wa wayy like by the method of finite differences. Let’s calculate the process of switching on a coil with a ferromagnetic core to the constant voltage. Let’s write the equation of the circuit d ri U . dt We consider that the dependence f(i) is giv given en (Fig. 8.2). We introduce the computational grid with a step h.

d dt

kk 1 tt

k

h

k

h

(8.3)

, then the equation (8.3) is represented as

follows: k

h

Ur

ik

or kk 1

hU

rik .

(8.4)

Let t 0t 0 be, at 0 0, 0i 0 (at the zero initial conditions). At t t1 0t h we have 1 h(U – ri0) according to the formula 0 (8.4). Knowing 1 with the help of the characteristic f( i) we find 1. i At 2 h(U – ri1), by 2 we find i2 etc. 1

117

8.3. The method of calculating transients in a nonlinear circuit based on the conditional linearization of the circuit equation If the equation of a nonlinear circuit contains a term, whose role is in significant in the transient, it is possible to use this term so that the nonlin ear equation becomes linear linear.. Let’s consider the process of switching on a coil with a ferromagnetic core to the sinusoidal voltage. In this case the equation of the circuit looks as follows: d ri U musin t . dt The dependence f (i) is shown in Fig. 8.5. Let the term ri have a secondary importance in d comparison with the term . In practice it takes dt powerful erful Fig. 8.5. The dependence a place at switching on non loaded pow f(i) transformers. Let’s write ri

r . Since this term is an in L

significant term (r is little), it is possible to admit that L we hav have e already a linear equation: dr dt

sin

Its solution is At

00

;

tA

mu

A

Utmusin

L

sin

mu

.

e

r t L

.

. The transient runs at

more intensiv intensively ely.. Let case A

Fig. 8.6. The dependence f(t)

118

const here. Then

m

and

u

u

2

cos t

mm

2

be, in this e

r t L

(Fig. 8.6). r Since is little, it is possible to reach the L maximal magnitude of flux linkage in a half period. It is almost equal to the double mag nitude of the peak m.

Knowing f( t) and f(i) it is possible to find the dependence i f(t) (Fig. 8.7). The initial peak of current will Fig. 8.7. The dependence f(t) lead to forming large electrodynam ic forces (these forces are proportion al to the square of current), which can cause a mechanical destruction of the winding. Therefore the non loaded pow powerful erful transformers are switched on through an additional resistor resistor,, which later is short circuited. r In the case of switching on a small inductiv inductive e coil, for which is big in L 1 d comparison with (T is the cy cycle cle of the applied voltage), the term T dt will be secondary secondary.. That’ That’s s why Li.. It is possible to consider that L is a Li di Lr : constant value. In this case we hav have e an equation equation: dt We solve this equation and get:

iI

sin t

mu

I

m

sin

u

i U musin t

e

r t L

.

.

Then we can find f(t) by using the curv curves es f(i) and i f(t). We consider that the examined method, if it is allowable, is the method, giving an approximate solution. In engineering practice such a solution can be useful.

8.4. The method of piecewise linear

approximation

Let’s consider the given method, using the example of calculating the switching on a coil with a ferromagnetic core to the con stant voltage (U 0) (Fig. 8.1). d ri U 0. dt Let’s approximate the curve f(i) by two tw o segments of a straight line (Fig. 8.8). On segment 1 the inductance is defined Fig. 8.8. The Weber Ampere characteristic as follows: 119

m L11

tan

mI

const,

m tan 2 const. mI The equation of the circuit is linear on each segment. The process on

and on segment 2 it is defined as follows 2 L

segment 1 0 tt

I0

is described by the equation:

di Lr10 dt

i U

U0

i

r

r t L1

1e

.

At the end of this segment we hav have e the current I which the timetI 0 , to

0

corresponds, i.e.

U0

r t L1 I 0

1e

r

The processes on segment tt2 tions: it

I0

I0

I 0.

have hav e the following initial condi

I 0 and described by the equation: r tt L2

U0

I0 di Lr20 i U i A e. dt r Let’s find the constant of integration taking into consideration that we

have hav eIA00

UU rr

A

I

We form the final solution:iI

at tt

UU rr

I0

.

0

e.

r tt L2

I

0

REFERENCES Main literature 1. Нейман Л.Р Л.Р.., Демирчян К.С К.С.. ТОЭ. — Л.: Энергоиздат Энергоиздат,, 1981. — Т. 1,2. Neumann L.R. L.R.,, Demirchan K.S. Theoretical fundamentals of elec trical engineering. — L.:Energoizdat, 1981. — V. 1, 2. 2. Теоретические основы электротехники / Под ред. П.А. Ион кина. Т. 1,2. — М.: ВШ., 1976. Theoretical fundamentals of electrical engi neering / Edited by P.A. Ionkin. — V. 1, 2. — M.: VS, 1976. 3. Теоретические основы электротехники / Под ред. Г.И. Ата бекова. Т. 1,2. — М.: Энергия, 1979. Theoretical fundamentals of electri cal engineering / Edited by by.. G.I.Atabeko G.I.Atabekovv. — V. 1, 2. — M.: Energia, 1979. 4. Сборник задач и упражнений по ТОЭ / Под ред. П.А. Ионкина. — М.: Энергоиздат Энергоиздат,, 1982. Collection of problems and exercises on theoretical fundamentals of electrical engineering / Edited by P.A.Ionkin. — M.: Ener goizdat, 1982. 5. Пашенцев И.Д. Методические пособия по решению задач курса ТОЭ. — Л.: ЛИИЖТ ЛИИЖТ,, 1981. — Ч. I—VI. Pashenchev I.D. Methodical ap pliances on solving problems of the course theoretical fundamentals of elec trical engineering. — L.: LIIZT LIIZT,, 1981. — V. I—VI. 6. Бессонов Л.А. Сборник задач по ТОЭ. — М.: ВШ, 1988. Bessonov L.A. Collection of exercises on theoretical fundamentals of electrical engi neering. —M.: VS, 1988. 7. Бессонов Л.А. Теоретические основы электротехники. — М.: ВШ, 1978. — Т.1,2. Bessonov L.A. Theoretical fundamentals of electrical engi neering. — M.: VS, 1978. — V.1, 2. 8. Новгородцев А.Б А.Б.. 30 лекций по теории электрических цепей. — СПб.: Политехника, 1995. Novgorodchev A.B. 30 lectures on the theory of electrical circuits. — SPb.: Polytechnika, 1995. 9. Шимони К. Теоретическая электротехника. — М.: Мир, 1964. — 773 с. Shimoni K. Theoretical electrical engineering. — M.: Mir Mir,, 1964. 10. Основы теории цепей / Г.В. Зевеке, П.А. Ионкин, А.В. Не тушил, С.В. Страхов. — М.: Энергоиздат Энергоиздат,, 1989. A fundamentals of the theory of circuits / G.V G.V.. Zev Zeveke, eke, P.A. Ionkin, A.V A.V.. Netushil, S.V S.V.. Straxov Straxov.. — M.: Energoizdat, 1989. 121

11. Белецкий А.Ф. Теория линейных электрических цепей. — М.: Радио и связь, 1986. Belechki A.F. Theory of linear electrical circuits. — M.: Radio i svjaz, 1986. 12. Поливанов К.М. Теоретические основы электротехники. — М.: Энергия, 1975. — Т.3. Polivanov K.M. Theoretical fundamentals of elec trical engineering. — M.: Energia, 1975. — V.3. Literature of the informativ informative e methodical pro providing viding the teaching pro cess worked out by the department «Theoretical fundamentals of electrical engineering».

Additional literature 1. Матханов П.Н П.Н.. Основы анализа электрических цепей. — М.: ВШ, 1990. Matxanov P.N. Fundamentals of analysis of electrical circu its. — M.: VS, 1990. 2. Практикум по ТОЭ / Под ред. Шакирова М.А. — СПб.: СПбГТУ СПбГТУ,, 1995. — Ч. 1, 2, 3. Practical work on theoretical fundamentals of electrical engineering / Edited by Shakirov M.A. — SPb.: SPGTU, 1995. — V. 1, 2, 3. 3. Шебес М.Р М.Р.. Задачник про теории линейных электрических цепей. — М.: ВШ, 1973. Shebes M.P M.P.. Book on the theory of linear elec trical circuits. — M.: VS, 1973. 4. Демирчян К.С К.С., ., Бутырин П.Л. Моделирование и машинный расчет электрических цепей. — М.: ВШ, 1988. Demirchan K.S., Butirin P.L. Modelling and machine calculating electrical circuits. — M.: VS, 1988. 5. Кухаркин Е.С. Основы технической электродинамики. — М.: ВШ, 1969 — Ч. 1, 2. Kuxarkin E.S. Fundamentals of technical electrody namics. — M.: VS, 1969. — V. 1,2. 6. Sergio Franco Franco.. Electric Circuits Fundamentals. — San Francisco: OUP USA, 1995. 7. Howatson A.M A.M.. Electrical Circuits and Systems. — Univ University ersity of Ox ford, 1996. 8. Bobrow L. Elementary Linear Circuit Analysis. — Univ University ersity of Mas sachusetts: OUP USA, USA, 1987. 638 p. 9. Bobrow L. Fundamentals of Electrical Engineering. — University of Massachusetts: OUP USA, USA, 1996. The use of the programs demanding understanding physical phenome na while stating the research problem of electromagnetic processes and the least spending time to carry out its solution on computer computer,, including using the Pspice, Workbench, Matlab, Mathcad software packages are recom mended. 122

Appendix 1 THE HO HOMET METASK ASK «THE CALCULA LCULATIO TION N OF TRANSIENTS IN LINES WITH DIS DISTRIBUTED TRIBUTED PARAMETERS» The hometask is giv given en as a code of four numbers indicating the num bers of variants in the appropriate tables. Using the data of the code it is necessary to write a complete condition of the task due to the following form: the code of hometask: «10c, 5 i, 12, 23». Where «10» is the number of the circuit in Table A. A.1.1, 1.1, «c» is the number of the line in Table A. A.1.1; 1.1; «5» is the number of the circuit in Table. A. A.1.2, 1.2, «i» is the number of the line in Table A. A.1.2; 1.2; «12» is the number in Table. A. A.1.3; 1.3; «23» is the number of the line in Table A. A.1.4. 1.4. The system consists of tw two o homogeneous lines connected according to the circuit in Fig. A. A.1.1. 1.1. The constant voltage 0Uis switched on to termi nals 1 1, the voltage magnitude is given by the teacher teacher.. The parameters of elements of a tw two o port netw network, ork, which is at the position of joint of lines, are giv given en in Table A. A.1.1. 1.1. The parameters of elements of a tw two o terminal netw network ork at the end ter minals of the second line are giv given en in Table A. A.1.2. 1.2. The parameters of the line are determined by the data of the experi mental measures of the input impedances at the idling regime (Table A. A.1.3) 1.3) and the short circuited regime (Table A. A.1.4). 1.4).

Fig. A. A.1.1. 1.1. The system, consisting of tw two o homogeneous lines

Fig. A. A.1.2. 1.2. The system, consisting of tw two o homogeneous lines (example)

123

Fig. A. A.1.3. 1.3. The equivalent circuit

Fig. A. A.1.4. 1.4. The plots of distributing u 1, u 1, i 1, i 1, u 2, and i 2 along the line

You should find the formulas for voltage and current and to draw the plots of changing these values along the lines at the time mo moment ment when the wav aves es of voltage and current, reflected by the end terminals of the second line, will pass n % of the length of the second line (n % is given by the teacher). You should be sure that homogeneous lines are undistorting. 124

125

The variant

The parameters

The circuit

The variant

The parameters

The circuit

d

500

400

c

r0

80

300

20

d

150

b

50

c

40

100

10

b

300

r 03

а

100

a

r 01

0.385

0.417

0.455

0.556

F

C0

50

200

70

60

r 02

d

c

b

а

150

10

30

40

r 04

d

c

b

а

200

200

100

100

r 01

136

162

118

154

nH

L0

400

200

400

100

r0

25

100

150

25

r 01

30

200

80

60

r 02

0.171

0.150

0.154

0.125

H

L0

25

300

50

75

r 03

d

c

b

а

20

200

120

40

r04

2

100

400

300

200

r 01

2.22

1.25

1.67

F

C0

The parameters of a tw two o port network at the position of joint of lines

r0

d

c

b

а

400

100

100

400

300

0.65

0.60

0.55

0.45

1.250

1.715

1.835

1.625

F

C0

H

L0

100

500

400

r0

Table A. A.1.1 1.1

126 The variant

The parameters

The circuit

The variant

The parameters

The circuit

200 300 400

b c d

1200 400 100 100

а b c d

r 02

100

а

r 02

500

20

200

100

r01

100

400

400

100

r01

0.112

0.04

0.15

0.10

H

L0

0.311

0.309

d

c

b

а

d

c

b

100

200

100

1200

r 02

400

300

200

100

H 0.242 0.28 а

r 02

L0

500

300

20

100

r 01

100

200

400

100

r 01

4.45

2.4

6.25

2.5

F

C0

0.803

0.834

F 1.035 0.893

C0

d

c

b

а

d

c

b

а

500

300

20

100

r02

100

100

400

1200

r02

100

100

200

100

1200

r01

500

20

200

r01

0.04

0.15

0.10

H 2.5

4.45

2.4

6.25

L0

0.112

H

L0

Table A. A.1.1 1.1 (continuation continuation))

127

The variant

The parameters

The circuit

The variant

The parameters

The circuit

200 400 800

b c d

100 200 300 400

а b c d

r0

100

а

r0

0.7

0.6

0.8

1

C0 F

2.08

2.36

2.25

3

2.5 d

c

b

1.0

2.0

1.3

1.6

FH 1.5 2

L0 а

400 1.875

200

50

100

C0 F

C0

d

c

b

а

r0

d

c

b

а

d

c

b

а

L0 H

0.4

1.4

0.8

1.2

H 1.0

L0

800

200 0.25

100 0.24

400 0.35

r0

d

c

b

а

d

c

b

а

r0

1.06

1.2

1

1.8

1.5

2.0

F 2.5

C0 а

d

c

b

400 1.33

50

200

100

L0 H

d

c

b

400

300

200

100

C0

0.1

0.8

0.4

F 0.625

C0

а

r0

d

c

b

а

1.25

1.45

1.87

3.12

F

100

800

300

250

r0

d

c

b

а

r0

0.13

0.08

0.20

0.50

1.00

1.20

F 1.65

C0

400

300 0.171

200

100

L0 H

Table A. A.1.1 1.1 (ending ending))

The circuit of load The parameters The variant

128 7.00 6.00 3.00 3.51 4.00 4.50 1.25 1.60 2.80 2.50

a b c d e f g h i k

F

Cload

200

250

500

500

140

125

250

200

100

100

rload

k

i

h

g

f

e

d

c

b

a

1.80

2.00

1.40

1.50

1.30

1.80

1.00

0.80

1.20

1.00

F

Cload

200

100

400

300

200

100

400

300

200

100

428

500

572

694

750

1000

1330

1750

300

rload1 r load2

k

i

h

g

f

e

d

c

b

a

3.2

3.00

1.00

1.40

0.70

1.00

0.9

1.5

2.3

1.00

F

100

80

500

100

500

300

400

200

100

420

80

100

100

500

500

400

300

100

200

100

Cload rload2 r load1

g

k

i

h

f

e

d

c

b

a

2.86

4.00

3.08

3.10

1.67

1.31

1.90

3.00

3.04

75

60

80

90

100

300

100

75

90

80

110

70

90

80

300

100

200

100

80

100

50

700

800

500

200

200

200

300

450

800

r load3 r load2 r load1

2.94

F

Cload

The parameters of a tw two o terminal network at the end terminals of the second line

Table A. A.1.2 1.2

129

The circuit of load The parameters The variant

0.070 0.040 0.090 0.085 0.093 0.076 0.050

c d e f g h i 0.045

0.102

b

k

0.400

a

F

L load

150

118

214

295

310

300

100

200

400

400

rload

k

i

h

g

f

e

d

c

b

a

400

400

300

300

200

200

100

100

100

100

200

100

400

300

200

100

400

300

200

100

0.0800

0.0445

0.1270

0.1050

0.0750

0.0428

0.1110

0.0938

0.0714

0.0417

HH

rload1 r load 2 L load

k

i

h

g

f

e

d

c

b

a

0.18

0.22

0.15

0.8

0.14

0.191

0.123

0.14

0.25

0.3

300

200

400

100

300

100

200

100

400

300

200

300

100

400

100

300

100

200

300

400

L load rload1 r load 2

k

i

h

g

f

e

d

c

b

a

0.081

0.081

0.150

0.193

0.133

0.083

0.085

0.087

0.063

0.063

H

90

80

75

70

60

80

90

100

300

100

80

100

110

60

70

90

80

300

100

200

500

200

200

200

300

800

500

700

600

800

L load r load 3 r load 2 r load1

Table A. A.1.2 1.2 (ending ending))

Table A. A.1.3 1.3 The data of measuring impedances of the idling regime and the short circuited regime of the first line (100 km km)) The variant

f

Z i.r i.r..1

Z s.c. s.c.1 1

1

Hz 50

437exp(–jj2.60) 437exp(–

280exp(jj2.60) 280exp(

2

50

583exp(–jj6030') 583exp(–

210exp(jj6.50) 210exp(

3

105

500exp(–jj60) 500exp(–

320exp(jj60) 320exp(

4

98

657exp(–jj14040') 657exp(–

221exp(jj14.60) 221exp(

5

200

500exp(–jj14.50) 500exp(–

320exp(jj14.50) 320exp(

6

202

666exp(–jj370) 666exp(–

240exp(jj370) 240exp(

7

300

602exp(–jj35.50) 602exp(–

415exp(jj35.50) 415exp(

8

400

385exp(–jj200) 385exp(–

416exp(jj200) 416exp(

9

400

380exp(–jj280) 380exp(–

420exp(jj280) 420exp(

10

485

320exp(–jj280) 320exp(–

500exp(jj280) 500exp(

11

490

301exp(–jj300) 301exp(–

528exp(jj300) 528exp(

12

800

400exp(+jj30) 400exp(+

625exp(–jj30) 625exp(–

13

800

0) 300exp(jj6.5 300exp(

835exp(–jj6.50) 835exp(–

14

1600

0) 657exp(–jj14.6 657exp(–

221exp(jj14.60) 221exp(

15

1700

030') 500exp(–jj14 500exp(–

320exp(jj14.50) 320exp(

16

1800

0) 602exp(–jj35.5 602exp(–

415exp(jj35.50) 415exp(

17

2280

0) 300exp(jj6.5 300exp(

835exp(–jj6.50) 835exp(–

18

3090

0) 657exp(–jj14.6 657exp(–

221exp(jj14.60) 221exp(

19

3200

0) 500exp(–jj14.5 500exp(–

320exp(jj14.50) 320exp(

20

3300

030') 602exp(–jj35 602exp(–

415exp(jj35.50) 415exp(

21

3750

0) 300exp(–jj6.5 300exp(–

835exp(–jj6.50) 835exp(–

22

4580

0) 657exp(–jj14.6 657exp(–

221exp(jj14.60) 221exp(

23

4700

030') 500exp(–jj14 500exp(–

320exp(jj14.50) 320exp(

24

4780

0) 602exp(–jj35.5 602exp(–

415exp(jj35.50) 415exp(

25

6075

0) 657exp(–jj14.6 657exp(–

221exp(jj14.60) 221exp(

26

6200

0) 500exp(–jj14.5 500exp(–

320exp(jj14.50) 320exp(

27

6580

0) 602exp(–jj35.5 602exp(–

415exp(jj35.50) 415exp(

130

Table A. A.1.4 1.4 The data of measuring impedances Z i.r s.c.2 of the second line (50 m) i.r..2 и Zs.c.2 The variant

f, Hz

Z i.r.2 ,

Z s.c.2,

The variant

f, Hz

Z i.r.2 ,

Z s.c.2,

1

50

–j7630

j20.95

15

1400

–j42

j3807

2

100

–j3807

j42

16

1600

j42

–j3807

3

200

–j1882

j85

17

1700

j85

–j1882

4

300

–j1231

j130

18

1800

j130

–j1231

5

400

–j898

j178

19

1900

j178

–j898

6

500

–j694

j230

20

2000

j230.7

–j694

7

600

–j550

j290.3

21

2100

j290.3

–j550

8

700

–j444

j360

22

2200

j360

–j444

9

800

–j360

j444

23

2300

j444

–j360

10

900

–j290.3

j550

24

2500

j694

–j230.7

11

1000

–j230.7

j694

25

2600

j898

–j178

12

1100

–j178

j898

26

2700

j1231

–j130

13

1200

–j130

j1231

27

2800

j1882

–j85

14

1300

–j85

j1882

The example of calculation. Two homogeneous undistorting lines with the A.1.2) 1.2) lengths l1 and l2 and the wav wave e resistances1 Z z1 and Z2 z2 (Fig. A. are switched on to the source of the constant voltage U. In the position of joint of the lines the capacitance 0Cand the resistance0rare connected. There is a load at the end of the second line. The load parameters load1,r r load 2, L load are known before. The attenuation coefficients 1, 2, the phase coefficients 1, 2 and the speeds of wav waves es1 v 2v v are also given for each line. Define the law of distributing voltage and current along the lines at the time moment when the waves, reflected by the end terminals of the second line, hav have e passed half the second line. 1) The time interval, when the wav ave e front is moving along the first line from the beginning up to its end, is equal1 t l1/v. 2) The time interval, when the wav ave e front is moving along the second line from the beginning up to its end, is equal2 t 2l /v. 131

3) The time interval, when the wav ave e front reflected by the end termi nals, passes half the length of the second line, is equal t2t /v. 3 4) The front of wav waves es of voltage and current, reflected by the end termi nals of the first line, according to the conditions of the hometask, passes the wa way yl t v during the interval of time 2t 3t . t 5) Let’s write the equations for the voltages and currents through the lines, marking direct, reflected and refracted wav waves es by indexes 1, 1 and 2, ac cordingly,, in the position of joint of lines: uu cordingly

ii

11

1

i

As ii 11

i C i 2, where ii 22 uu11 , zz 11

,

uu 2

u , thati 1

1

2

u z1

1

u

11

1

uC

r0i 2 u 2,

i z . 2 2

i C i 2 ,u

1

u

1

r i0 2

z,22 iu u 1 i 1z1. 1 Sum up the tw two o last equations. 2zui11

1

r0i 2 i 2z 2 i 1z 1 i 2r

0

z 2.

The circuit, shown in Fig. A. A.1.3, 1.3, satisfies the last equation. This calcu lation circuit is fair only to define the law of changing voltages and currents as time functions at the end terminals of the first line (points 1 —1 ), as it has been obtained with the help of the equations written for these end points of the line. It is visible from the calculation circuit that the load, connected at the end of the second line, influences neither on the structure of the calcula tion circuit nor on the distribution of currents and voltages. It is explained that until the wav waves es reach the end of the second line and, having been re flected by this end, come to the position of joint of lines, the distribution of currents and voltages in the first line does not depend on the load at the end of the second line. 6) Having calculated the circuit (Fig. A. A.1.3) 1.3) we shall find the law of changing currents and voltages in the position of joint of lines as time func tions. The time t can be read off either fro from m the moment of coming the direct wa wave ve to the position of joint of lines (the mo moment ment of turning on the switch in the calculation circuit in Fig. A. A.1.3), 1.3), or from the moment of switching on the line under the action of the voltage U. In the last case the turning on the switch in the calculation circuit occurs at the moment 1. t 132

At calculating the circuit in Fig. A. A.1.3 1.3 it is necessary to take the magni tude of u 1 at the points 1 —1 (at the end point of the first line). If the loss free line is considered, then u1 U. We consider switching on the undis torting line with losses under the constant voltage U const therefore u 1 U exp(– 1x) in any point of the line where x is the distance read off from the beginning of the first line. Hence, at the end points 1 —1 of the first line u 1 U exp(– 1l 1) and we should consider 2 u1 2U exp(– 1l 1) at calculating the circuit. The calculation of currents and voltages can be made by any of the known methods of calculating transients in the linear circuits with the lumped parameters. Let’s calculate the circuit in Fig. A. A.1.3 1.3 by the classical method. Let’s write the equations of the circuit: du z11 iu CC 2u 1, u i 2 r 0 z2 , i 1 i 2 i C i 2 C 0 C . dt We express the current1,i through the voltage C u in the first equation. ud u duC zz r0 12 z10 CC Cu 2 u , z C u 2u 1. CC 1 1 0 dt rd zz t r 02 02 It is possible to write the solution of the last equation as: 2z ur10 2 uu u Aexp t , CC C r012 zz zz 12

r0

zz Cr 10

1 —1 C, (u0)

0 andA

;

0

2

. zz Cr z z r 10 0 2 1 2 0 The constant A is calculated by the initial conditions. If we make the time reading off from the moment of switching on the line, the turning on of the switch in the calculation circuit (at the moment of coming the wave to the end terminals 1 —1 of the first line) will occur at t 1t , and, hence, the initial condition will be uC(t 1) 0. We shall make the time reading off from the mo moment ment of coming the wave u 1 to the end terminals 1 —1 of the first line, i.e. from the moment of turning on the switch in the circuit. If we consider that there was no voltage in the second line at the moment of co coming ming the wav wave e to the terminals

where;

2z ur10 r012 zz

2

at t

0. 133

Whence voltages and currents as the time functions for the position of joint of homogeneous lines will be:

2z ur10

utC

it22

2

r012 zz 22 uu11

uC rr02 zz

0

z2

1

1e xp

u11 u

1e xp

i

, u 21

2zrt02

exp

2u it 1

u1 u

, i

rr02 zz

1

r012zz

1

1

uu11 zz 11 1

2z1

, i

02

z

z1

r 0 z1 z 2

r0

r012 zz

1

z2

, u 2 i 2 z2 u

1e xp tu

u 1,

z2

z1

,

1

exp

.

7) Let’s define the law of changing currents and voltages of reflected wav aves es and refracted wav waves. es. When there are no losses in the line, the wav waves, es, reflected by the place 1 —1 , and the refracted waves will co come me to any point of the line, at the distance x from the position of joint with a time dela delayy x/v. Hence, if you want to define the law of changing currents and voltages for reflected and refracted wav waves es at any point, distanced on x from the po sition of joint, it is necessary to replace t by t–x/v in the equations of the appropriate values, got for the position of joint of lines: u ur 10

iu 12

1

zz2

r012zz

uu11 , zz 10

2z r

1

2 1

z2

2 r 0 z 2 ex p

1 exp

tx

/v

tx

/v

,

u , i

2

2

z2

.

8) If the line with losses is considered, the equations for voltages and currents will be as follows: 134

u ur 10

1

zz2

r012zz

2z u ux22

2 r 0 z2 exp

1

12

tx

1e xp

r012 zz

u i

1

1

z1 u

i

2

2

z2

tx

/v

/v

exp

exp

x (,A. A.1.1) 1.1)

1

,

(A. A.1.2) 1.2)

,

(A. A.1.3) 1.3)

.

(A. A.1.4) 1.4)

9) Let’s determine u 1 and i 1 in each point of the first line for the moment t t . The equations (A. A.1.1) 1.1) and (A. A.1.3) 1.3) are fair for any point of the line at x t v. In the points x t v only direct wav aves es1 and of u i 1 take u

place: u 1

Uexp(– 1x), i

1

1

z1

.

10) Let’s draw the plots of distribution of u1, u 1, i 1, i 1, u 2 and i 2 along the line for the considered case (Fig. A. A.1.4). 1.4). 11) Let’s calculate the distribution of voltage u2 and current i 2 along the second line using the equations (A. A.1.2) 1.2) and (A. A.1.4) 1.4) for the time mo ment t t too. The reflection of wa waves ves fro from m the end terminals of the second line 2 —2 can be considered similarly as the previous case by constructing the calcu lation circuit for the end terminals 2 —2 and determining the functional dependences of voltages and currents on time for the end points of the second line 2 —2 with the help of the calculation circuit. For the voltage across the end terminals 2 —2 of the circuit u load and the current iload through it we shall get the following equations:

uu 22 ii22 where urload

i load12

L load

u i

diLL ; L dt

uload,

2

(A. A.1.5) 1.5)

i load ,

2

di load

dt

r

(A. A.1.6) 1.6) i .

load2

135

A.1.6) 1.6) and sum Expressing i 2 and i 2 by u 2 and u 2 in the equation (A. ming up the equations (A. A.1.5) 1.5) and (A. A.1.6), 1.6), we shall get: uu22

uload

uu22

i2z 2

2z ui 22

2

uload .

From the last dependence it is visible that we can define 2 asi a current, arising in the calculation circuit in Fig. A. A.1.5, 1.5, switched on under the volt active e resistance equal to the wav wave e age 2u 2. This circuit consists of the activ resistance 2z of the second line, and the load, connected with this resis tance in series.

Fig. A. A.1.5. 1.5. The calculation circuit

At calculating the considered circuit it is necessary to use the magni tude of u 2 at the points 2 —2 (at the end points of the second line). In the received ed from the equation considered case the magnitude of 2ucan be receiv (A. A.1.2) 1.2) and it will be equal 2z u 12 tl 2 / v ul 22 1e xp exp , 2 zz r 12 0 if the time t is read off from the moment of the beginning of mo moving ving re fracted wa wav ves from the position of joint of tw two o lines. Let’s read off the time from the moment of co coming ming refracted wa wavves to the end terminals 2 —2 of the second line, i.e. t 0 is the moment of 136

turning on the switch in the calculation circuit. Then in the points 2 —2 will look as: 2z u ul 22

12

zz 12

r0

1 exp

tD

exp

2

2

1 exp

t

. (A. A.1.7) 1.7)

As the refracted wav wave e u2 is the time function, we shall use Duamel’s integral in the form: t

it

u 0.Y t

∫u

yY t

yd y

0

to calculate the last circuit, switched on under the action of the voltage 2u 2, then t

it22

20 u

Yt2

∫ 2u

2

y Yt2

y dy.

(A. A.1.8) 1.8)

0

It is also possible to use the other known methods of calculating tran sients. For using Duamel’s integral it is necessary to calculate the transient admittance Y2 (t) of the calculation circuit. For it we shall consider the switching on this circuit under the constant voltage U. We shall calculate 2Y(t) by the operator method. AsiI 00 , p L 2 Up . Zp The equivalent operator impedance of circuit shown in Fig. A. A.1.6 1.6 is equal Zp rrload21

rload12 z load

zz2

rp L load2 rp load2

pL r load1 r

rp load2

L

L load2

2

.

By substituting the equation of Z(p) into the equation for2(Ip), we shall get: Ur load 2 pL Ip2 pr load21rload zz2 pL r load1 r load2 2 137

Fig. A. A.1.6. 1.6. The calculation circuit

Urload2 pL rrload12 rrload 21

where

Lr

rload12 z rload12 z

UL

zz2 p

load

z2

load

r load12 load

z2

L rrload12

load

2

,

p

.

Then it2

UU rrload12 zz

1e xp

t load1

U rrload12 zz

r load2

rload 2

1e load1

xp

r load2

e xp

t

2

t

.

2

From this equation we shall get

Yt2 where AB

U rrload12 zz

it2 U

A 1e B xp rload2

; load1

r load2

t ,

. 2

Let’s determine the equations of the other values of the formula (A. A.1.8). 1.8). From the formula (A. A.1.7): 1.7):

Yt22 138

y

A 1e B xp

t

y

; u

0

0;

4z ul 22 exp

2uy2

zz 12

yy

1 exp

r0

4z u

Dy

2x uy22

22

p

; D

D 1 exp

12

zz 12

r0

exp

;

l2 .

Let’s determine the current as the time function at the end terminals 2 —2 of the second line (the current through the load) according to the formula (A. A.1.8): 1.8): t

ii2

load

∫

Dy exp

A 1 B exp

t exp y dy

0 tt

ADB

exp

∫∫

1

ty exp

AD

dy

y

exp

dy

00

AD

exp

t

t

ADB 1

0

AD ADexp

AD 11

yt

ADB 1

exp

exp y

0

1

exp

Bt exp 11

AD 1e M xp

t

t

B

t

exp ex p

exp

N exp

t

t

t

.

12) Let’s calculate the reflected wav waves es of voltage and current through the second line. Using the conditions for voltages and currents at the end terminals of the second line, we shall get:

uu22

i 2z 2

1 exp

tt

AD 1 M exp u

i

2

2

z2

N exp t

z ;

2

. 139

In order to get the equations for u2 and i 2 in any point of the second line, at the distance x 3vt from the end terminals of the second line at the time moment t3 it is necessary to replace t by t /v in the equations of 3–x u 2 and i 2, receiv received ed for the end terminals, and to multiply these equa tions (for the line with losses) by exp( 2x). Then

uA2

tx D 1e xp 33 2

/v

D 1 M exp

/ v

tx u

Ntexp

32

x / v z exp

2

x , i

2

2

z2

.

The last equations are fair for the points where x3v, tas there are no reflected wa waves ves in the points where x 3v,t and there are only refracted wav aves es of u 2 and i 2. 13) Let’s draw the plots of distributing wav waves es of voltage and current along the lines for the considered example in a case: 400 , 1 1 zz2 3 3 0.82·10 1/ 1/km, km, 2 3·10 1/ 1/km, km, r 0 100 , C 3.6 F, load r 1 0 600 , load r 2 1000 , Lload 2H, 1l 800 km, 2l 400 km, v 3 A.1.7). 1.7). 105 km/sec (Fig. A.

Fig. A. A.1.7. 1.7. The plots of distributing wa waves ves of voltage and current along the lines

140

Appendix 2 THE HO HOMET METASK ASK «THE CA CALCULATI LCULATIO ON OF A NONLINEAR MA MAGNETIC GNETIC CIR CIRCUIT» CUIT» The magnetic circuit is giv given en in Fig. A. A.2.1. 2.1. The magnetic core is as sembled with sheets made of electrotechnical steel. The normal magne tization curv curve e of steel is approximated by the equation H 1b a+a2b3. The thickness of a sheet is equal 2d 0.15 mm mm.. The voltage mucos U t the is applied to the winding. The magnitudes of the voltage peak m, U frequency f, the parameters of the magnetic core and the number of turns w of the winding are giv given en in Table A. A.2.1. 2.1. The factors a a2 and Sh 1 and teinmest’s factor , by which we calculate hysteresis losses, are giv given en in Table A. A.2.2. 2.2. It is required: 1. To define the laws of changing the magnetic flux and induction in the magnetic core in time, neglecting the activ active e resistance of the winding wire, the stra stray y fluxes and the swelling fluxes in the air gaps of the mag netic core. 2. To write the equation of the instantaneous value of current through the winding and to define its effectiv effective e value.

Fig. A. A.2.1. 2.1. The magnetic circuit

141

3. To calculate an activ active e pow power er,, caused by losses in the magnetic core, and also the apparent pow power er,, consumed from the source. 4. To construct the plots of the instantaneous values of current through the winding and the magnetic flux in the magnetic core. 5. To calculate the effectiv effective e value of electro electromagnetic magnetic forces, compress ing non metal gaskets in the air gaps of the magnetic core. The task of a higher difficulty. To define at what magnitude of a direct current through the winding the forces, compressing gaskets in the air gaps will be the same. The numbers of a variant fro from m Table A. A.2.1, 2.1, Table A. A.2.2, 2.2, are giv given en to a student by the teacher teacher.. The notes: S1 S3 S2/2; l 1 3l as 2. Choose the right identically ratio of their parameters. The answ answers ers are: a)1Z> Z2, b) Z1 < Z 2, c) 1 > 2, d) 1 < 2, e) 1 > > 2, f) 1 < 2, c) 1 > 2, c) 1 < 2. 10. In which of the ranges: a) 0.25 < l h) is in the homogeneous alternating electro electromagnetic magnetic field, the lines of the magnetic induction are directed along the sheet perpendic ularly to its cross section. Where will the specific pow power er of losses be more: а) in the centre of the sheet; b) near to the surface; с) identical ev everywhere. erywhere. The answ answers ers are: а), b), с). 22. How does the phase shift angle betw between een the equivalent sine wav aves es of current and voltage of a coil with a ferro ferromagnetic magnetic core change when we use material with the smaller area of a hysteresis loop? а) will increase; b) will decrease; с) will not change. The answ answers ers are: а), b), с). 23. How does the phase shift angle betw between een the equivalent sine wav aves es of current and voltage of a coil with a ferro ferromagnetic magnetic core change when we increase the thickness of the sheets of the core? а) will increase; b) will decrease; с) will not change. The answ answers ers are: а), b), с). 24. Does the complex magnetic resistance of a magnetic core depend on the method of choosing the equivalent sine wav aves es of current and volt age? а) Yes; b) No. The answ answers ers are: а), b). 25. Which order has the system of algebraic equations of the method of harmonic balance at calculating k harmonics in the common case? а) k; b) 2k; с) k–1. The answ answers ers are: а), b), с). 26. Is the statement fair that the calculation by the method of a harmonic balance demands to use the method of superposition? а) Yes; b) No. The answ answers ers are: а), b). 27. Is it necessary for analysing the stability of a balance state to giv give ea deviation: а) by current; b) by voltage? The answ answers ers are: а), b). 28. Which of the written equations is used at researching the phenom phenome e na in an electrical circuit having a coil with a ferromagnetic core, which is fed by a sine wav e voltage? di d . The answ answers ers are: а)uR iL ; b) uR iw dt dt 29. Which of the giv given en equations corresponds to the instantaneous val ue of an EMF EMF,, arising in a coil with a ferromagnetic core having 100 turns if the magnetic flux 0.05sin(( t 0.05sin /6); where 314 radian/sec.? 194

The answ ers are: а) 1570sin ( t 3 / 2); b) 1570sin ( t 3 / 2); с) 1570sin ( t /2); d) 1570sin ( t /2). 30. Which of the giv given en equations corresponds to the instantaneous val ue of a magnetic flux of a coil with a ferromagnetic core if it has 100 turns and it is supplied by the voltage u 3140 3140ssin t at a frequency of 50 Hz Hz? ? The answ answers ers are: а) 0.01sin( t /2); b) 0.01sin( t /2); с) 0.1sin( t /2); d) 0.1sin( t /2). 31. Which of the listed high harmonics are in the curv curve e of current, run ning through the coil with a ferromagnetic core, which is supplied by a sine wav e voltage? The answ answers ers are: а) ev even; en; b) odd; с) ev even en and odd; d) zero and ev even. en. 32. In which of the indicated dependences is there a wattless co compo mpo nent of current, running through the coil with a ferromagnetic core, and a frequency of current? The answ answers ers are: а) a squared dependance; b) an inv inverse erse squared de pendance; с) a directly proportional dependance; d) an inv inversaly ersaly propor tional dependence. 33. Which of the listed factors determines basically the form of mag netic lines in any magnetic circuit? The answ answers ers are: а) air gaps, b) ferromagnetic cores; с) windings; d) excitating currents.

A.4.15. A. 4.15. The autooscillations in nonlinear electrical cir circuits cuits 1. Is it possible that autooscillations arise in the electrical circuit in which а) a positiv positive e feedback or b) a negativ negative e feedback is made? The answ answers ers are: а), b). 2. Is it possible that undamping oscillations arise in the passiv passive e circuit containing: а) nonideal coils of inductance; b) nonideal capacitors; с) os cillations are impossible? The answ answers ers are: а), b), с).

A.4.16. A. 4.16. The tr transients ansients in nonlinear electrical cir circuits cuits 1. What is the adv advantage antage of the analytical solution of a task when we calculate a transient in the nonlinear electrical circuits: а) in the simplicity of solution; b) in the accuracy of solution; с) in the opportunity to deter mine the influence of the circuit parameters on the kind of a transient more simpler? The answ answers ers are: а), b), с). 195

2. We approximate the characteristic of a nonlinear element by line seg ments. What conditions should be accepted at the borders of the segments to find the integrating constants if the differential equation has the first order on each segment: а) the continuity of a decision variable at transfer ring from one segment to the next; b) the continuity of a decision variable and its kth derivativ derivative; e; с) the continuity of a decision variable and its (k– th 1) derivativ derivative. e. 3. Which of the listed methods allows to calculate a transient in the nonlinear electrical circuit more simpler? The answ answers ers are: а) the method of conditional linearization; b) the meth od of piecewise linear approximation; с) the step by step method; d) the graphic method. 4. How can we increase the accuracy of calculating a transient by the method of piecewise linear approximation? The answ answers ers are: а) by considering the initial conditions more exactly exactly,, b) by increasing the number of the segments of the approximating broken line, с) by the linearization of the characteristic of a nonlinear element. 5. The parameter (R, L, C) changes by a jump at the piecewise linear approximation of the characteristic of a nonlinear element at the borders of the segments. Does the transient arise in this case? The answ answers ers are: а) Yes; b) No; с) in the dependence on which of the parameters changes. 6. The appropriate magnitude (R, L, C) changes by a jump at the piece wise linear approximation of the characteristic of a nonlinear element at the borders of the segments. Is it necessary to consider the instantaneous changing of the parameter at calculating the transient on each segment? А) Yes; b) No. The answ answers ers are: а), b).

The answ answers ers A.4.1. The main conceptions. The laws of electrical circuits A.4.1. 1. c); 2. a); 3. b); 4. d); 5. a); 6. a); 7. a); 8. c). A.4.2. A. 4.2. The properties and methods of calculating linear circuits of a direct current 1. b); 2. b); 3. c); 4. a); 5. a); 6. c); 7. d); 8. d); 9. a); 10. c); 11. a); 12. d); 13. d); 14. c); 15. b). A.4.3. A. 4.3. The properties and methods of calculating circuits of a sine wa ve current 1. b); 2. c); 3. a); 4. d); 5. a); 6. a); 7. c); 8. d); 9. a). A.4.4. A. 4.4. The fundamentals of the complex method of calculating circuits of a sine wave current 1. a); 2. b); 3. a). A.4.5. A. 4.5. The inductiv inductively ely coupled circuits 1. c); 2. b); 3. b); 4. b); 5. b). A.4.6. A. 4.6. The three phase circuits 1. d); 2. b); 3. b); 4. d); 5. b); 6. d); 7. c); 8. d); 9. a); 10. a). A.4.7. A. 4.7. The linear electrical circuits of a non sine wa ve current 1. c); 2. b); 3. d); 4. a); 5. a); 6. a); 7. d); 8. a); 9. c); 10. b); 11. d); 12. b); 13. b), d). A.4.8. A. 4.8. The resonance phenomena in the electrical circuits and the frequency char acteristics 1. b); 2. c); 3. a); 4. d); 5. b); 6. c); 7. d); 8. b); 9. c); 10. a). A.4.9. A. 4.9. The tw two o port networks 1. c); 2. a); 3. c); 4. b); 5. c); 6. b); 7. с); 8. a),b); 9. c); 10. c); 11. a); 12. a); 13. b); 14. c); 15. c); 16. a). A.4.10. A. 4.10. The transients in linear circuits 1. d); 2. c); 3. b); 4. d); 5. a); 5. a); 6. c); 7. d); 8. c); 9. a); 10. b) ;11. c); 12. b); 13. b); 14. b); 15. a); 16. a); 17. At the time moment of acting a pulse — a), after ending a pulse b); 18. c); 19. [a), b), с)]. A.4.11. A. 4.11. The elements of synthesis of linear circuits with the lumped parameters 1. d); 2. [b), c), a), f), g), d), e), i), j), h)]; 3. [e), d), b), c), a)]; 4. a); 5. b); 6. [c), a) b)]. A.4.12. A. 4.12. The circuits with the distributed parameters in the steady regimes 1. a); 2. a); 3. a); 4. b),а),с); 5. [1 b) ; 2 b) ; 3 d) ; 4 b) ; 5 d) ; 6 d) , 7 g), 8 d) ; 9 e) ; 10 а); 11 с) ; 12 а)]; 6. b); 7. a); 8. b); 9. [b) , d) , g)]; 10. a); 11. b); 12. b); 13. a); 14. a); 15. b); 16. a); 17. b). A.4.13. A. 4.13. The transients in circuits with the distributed parameters 1. [1 b), 2 c), 3 a), 4 e), 5 d)]; 2. b); 3. [a), b) , с)]; 4. [a), b)]; 5. a); 6. d); 7. c); 8. a); 9. b); 10. a). A.4.14. A. 4.14. The nonlinear electrical and magnetic circuits 1. b); 2. c); 3. c); 4. a); 5. C; 6. [0, A, B]; 7. D; 8. [0, A, B, C, D]; 9. A; 10. B; 11. [C, D]; 12. b); 13. c); 14. b); 15. a); 16. b); 17. [a), b), c), d)]; 18. b); 19. a); 20. b); 21. b); 22. a); 23. b); 24. a); 25. b); 26. b); 27. [either a) or b)]; 28. b); 29. d); 30. c); 31. b); 32. d); 33. b). A.4.15. A. 4.15. The autooscillations in nonlinear electrical circuits 1. [a) and b)]; 2. b). A.4.16. A. 4.16. The transients in nonlinear electrical circuits 1. c); 2. a); 3. a); 4. b); 5 b); 6 b).

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CONTENTS CO NTENTS 1. SYNTHESIS OF LINEAR PASSIVE ELECTRI ELECTRICA CAL L CIR CIRCUITS CUITS WITH LUMPED PARAMETERS .................................................................................... 4 1.1. The properties of input functions of passive electrical circuits (tw two o pole networks) netw orks) ......................................................................................................... 5 1.2. The representation of input functions of tw two o pole networks as simple fractions (Foster’s method) ............................................................................. 7 1.3. The realization of input functions of a tw two o pole netw network ork with real and imaginary roots of the denominator (Foster’s method) ................................... 8 1.4. The realization of input functions of a tw two o pole netw network ork having only imaginary roots of the denominator .............................................................. 14 1.5. The realization of the input function having complex roots of the denominator .................................................................................................. 19 2. ELECTRIC ELECTRICAL AL CIRCUITS WITH DIS DISTRIBUTED TRIBUTED PARAMETERS (THE STEADY REGIME) ................................................................................... 23 2.1. Electrical circuits with distributed parameters .............................................. 23 2.2. The equations of a line with distributed parameters ...................................... 23 2.3. Solving equations of an uniform line (the steady sine wa wave ve regime) ............. 25 2.4. Running wav wave es ............................................................................................... 27 2.5. Characteristics of a uniform line. Conditions for an undistorting line .......... 30 2.6. An uniform line at various working regimes .................................................. 31 2.7. Working regimes of a loss free line ................................................................ 33 3. TRANSIENTS IN CIRCUITS WITH DIS DISTRIBUTED TRIBUTED PARAMETERS ........... 38 3.1. Transients in an uniform undistorting line (the classical method of calculation) .................................................................................................... 38 3.2. Transients in an uniform undistorting line (the operational method of calculation) .................................................................................................... 40 3.3. Refraction and reflection of wav waves es at the position of joint of tw two o uniform lines ............................................................................................................... 43 3.4. Reflection of wav waves es from the end of a line ..................................................... 45 3.5. Switching on an uniform line ....................................................................... 47 3.6. A case when there is an inductiv inductive e coil at the position of joint of tw two o uniform lines .................................................................................................. 49 3.7. A case when there is an activ active e resistance at the position of joint of tw two o uniform lines .................................................................................................. 51 4. NONLINEAR CIR CIRCUITS CUITS OF A DIRECT CURRENT ....................................... 53 4.1. Nonlinear elements and their characteristics ................................................ 53 4.2. Current voltage characteristics of some nonlinear elements ........................ 55

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4.3. The calculation of simple circuits with passiv passive e nonlinear elements .............. 58 4.4. The calculation of simple nonlinear circuits of a direct current by the iterativ iterative e method ....................................................................................... 61 5. LINEAR AND NONLINEAR MA MAGNETIC GNETIC CIRCUITS OF A DIRECT CURRENT ........................................................................................................... 66 5.1. The connection betw between een a magnetic field and an electrical current. Biot Savart Laplac’s law law.. Ampere’s circuital law .......................................... 66 5.2. Laws and parameters of magnetic circuits ..................................................... 68 5.3. The characteristics of magnetizing ferromagnetics ....................................... 71 5.4. Calculation of nonlinear magnetic circuits ................................................... 74 5.5. About the calculation of a permanent magnet .............................................. 77 6. NONLINEAR ELECTRI ELECTRICA CAL L AND MAGNETIC CIR CIRCUITS CUITS IN PERIODIC PROCESSES PR OCESSES ....................................................................................................... 79 6.1. Features of periodic processes in nonlinear circuits ...................................... 79 6.2. The method of equivalent sinusoids .............................................................. 81 6.3. Losses in a ferromagnetic core at the periodic changing of the magnetic flux ................................................................................................................. 82 6.4. The equation and the equivalent circuit of a coil with a ferromagnetic core ... 86 6.5. Complex reluctance ...................................................................................... 87 6.6. Ferroresonance at series connecting a coil with a ferromagnetic core and a capacitor .............................................................................................. 88 6.7. Ferroresonance at connecting a coil with a ferromagnetic core and a capacitor in parallel .................................................................................... 91 6.8. Controlled inductiv inductive e elements. The ferromagnetic amplifier of pow power er ......... 93 7. OSCILA OSCILATIO TIONS NS IN NONLINEAR CIR CIRCUITS CUITS ................................................... 96 7.1. The theory of stability of regimes in nonlinear electrical circuits ................. 96 7.2. Autooscillation circuits ................................................................................. 98 7.2.1. General conceptions ............................................................................... 98 7.2.2. Theory of active oscillators with an external feedback ......................... 102 7.2.3. Activ Active e oscillators with an internal feedback .......................................... 111 7.3. The relaxation oscillations .......................................................................... 114 8. CA CALCULAT LCULATIO ION N OF TRANSIENTS IN NONLINEAR CIRCUITS ................ 115 8.1. The graphic method of calculating transients ............................................. 11 115 5 8.2. The method of step by step intervals (Eiler’s method) .............................. 117 8.3. The method of calculating transients in a nonlinear circuit based on the conditional linearization of the circuit equation .............................. 118 8.4. The method of piecewise linear approximation ......................................... 119 REFERENCES ...................................................................................................... 121 Appendix 1. The hometask «The calculation of transients in lines with distributed parameters» ...................................................................... 123 Appendix 2. The hometask «The calculation of a nonlinear magnetic circuit» ..... 141 Appendix 3. The hometask «The calculation of transients in the nonlinear electrical circuit» ............................................................................... 15 152 2

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